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

The variable is called a parameter and can take values ​​from "minus infinity" to "plus infinity". Consider, for example, a value and substitute it into both equations: ... Or humanly: "if x is equal to four, then y is equal to one." A point can be marked 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 "te" parameter. As for the "normal" function, for the American Indians of a parametrically defined function, all rights are also respected: you can plot a graph, find derivatives, etc. By the way, if there is a 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 to find the derivative parametric function there is a formula:

Find the derivative of the "game with respect to the te variable":

All the rules of differentiation and the table of derivatives are, of course, also valid for the letter, thus there is no novelty in the process of finding derivatives... Just mentally replace all the "x" in the table with the letter "te".

Find the derivative of "x with respect to the te variable":

Now it only remains to substitute the found derivatives into our formula:

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

As for the designations, in the formula, instead of writing, it could simply be written without a subscript, since this is the "usual" derivative "by x". But in the literature there is always a variant, so I will not deviate from the standard.

Example 6

We use the formula

In this case:

Thus:

A 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 considered example, when I found it, I expanded the parentheses under the root (although I could not do this). Chances are great that when substituted 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 parametrically defined function

This is an example for independent decision.

The article The simplest typical tasks with derivative we considered examples in which it was required to find the second derivative of a function. For a parametrically given function, you can also find the second derivative, and it is found by the following formula:. It is quite obvious that in order to find the second derivative, one 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 found derivatives into the formula. For the sake of simplification, we use the trigonometric formula:

I noticed that in the problem of finding the derivative of a parametric function, quite often, for simplification purposes, one has to use trigonometric formulas ... Remember them or keep them close at hand, and do not miss the opportunity to simplify each intermediate result and answers. What for? 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 take a 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 consolidate the material, I propose a couple more examples for an independent solution.

Example 9

Example 10

Find and for a function defined parametrically

Wish you success!

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

Solutions and Answers:

Example 3: Solution:






Thus:

The function can be set in several ways. It depends on the rule that is used to define it. The explicit form of setting the function has the form y = f (x). There are times when its description is impossible or inconvenient. If there is a set of pairs (x; y) that need to be calculated for the parameter t over the interval (a; b). To solve the system x = 3 cos t y = 3 sin t with 0 ≤ t< 2 π необходимо задавать окружность с центром координат с радиусом равным 3 .

Defining a parametric function

Hence we have that x = φ (t), y = ψ (t) are defined on for a value t ∈ (a; b) and have an inverse function t = Θ (x) for x = φ (t), then in question on specifying a parametric equation for a function of the form y = ψ (Θ (x)).

There are times when, in order to investigate a function, it is required to search for the derivative with respect to x. Consider the formula for the derivative of a parametrically given function of the form y x "= ψ" (t) φ "(t), let's talk about the derivative of the 2 and n-th order.

Derivation of the formula for the derivative of a parametrically given function

We have that x = φ (t), y = ψ (t), defined and differentiable for a value t ∈ a; b, where x t "= φ" (t) ≠ 0 and x = φ (t), then there is an inverse function of the form t = Θ (x).

To begin with, you should go from parametric to explicit assignment. To do this, you need to obtain a complex function of the form y = ψ (t) = ψ (Θ (x)), where there is an argument x.

Based on the rule for finding the derivative of a complex function, we obtain that y "x = ψ Θ (x) = ψ" Θ x · Θ "x.

This shows that t = Θ (x) and x = φ (t) are inverse functions from the formula of the inverse function Θ "(x) = 1 φ" (t), then y "x = ψ" Θ (x) · Θ "(x) = ψ" (t) φ "(t).

Let's move on to considering the solution of several examples using the table of derivatives according to the differentiation rule.

Example 1

Find the derivative for the function x = t 2 + 1 y = t.

Solution

By hypothesis, we have that φ (t) = t 2 + 1, ψ (t) = t, hence we obtain that φ "(t) = t 2 + 1", ψ "(t) = t" = 1. It is necessary to use the derived formula and write the answer in the form:

y "x = ψ" (t) φ "(t) = 1 2 t

Answer: y x "= 1 2 t x = t 2 + 1.

When working with the derivative of a function h, the parameter t specifies the expression of the argument x through the same parameter t, so as not to lose the connection between the values ​​of the derivative and the parametrically specified function with the argument to which these values ​​correspond.

To determine the second-order derivative of a parametrically given function, you need to use the formula for the first-order derivative on the resulting function, then we get that

y "" x = ψ "(t) φ" (t) "φ" (t) = ψ "" (t) φ "(t) - ψ" (t) φ "" (t) φ "( t) 2 φ "(t) = ψ" "(t) φ" (t) - ψ "(t) φ" "(t) φ" (t) 3.

Example 2

Find the 2nd and 2nd order derivatives of the given function x = cos (2 t) y = t 2.

Solution

By hypothesis, we obtain φ (t) = cos (2 t), ψ (t) = t 2.

Then after transformation

φ "(t) = cos (2 t)" = - sin (2 t) · 2 t "= - 2 sin (2 t) ψ (t) = t 2" = 2 t

Hence it follows that y x "= ψ" (t) φ "(t) = 2 t - 2 sin 2 t = - t sin (2 t).

We obtain that the form of the derivative of order 1 is x = cos (2 t) y x "= - t sin (2 t).

To solve it, you need to apply the second-order derivative formula. We get an expression of the form

yx "" = - t sin (2 t) φ "t = - t" sin (2 t) - t (sin (2 t)) "sin 2 (2 t) - 2 sin (2 t) = = 1 sin (2 t) - t cos (2 t) (2 t) "2 sin 3 (2 t) = sin (2 t) - 2 t cos (2 t) 2 sin 3 (2 t)

Then the specification of the second order derivative using the parametric function

x = cos (2 t) y x "" = sin (2 t) - 2 t cos (2 t) 2 sin 3 (2 t)

A similar solution can be solved by another method. Then

φ "t = (cos (2 t))" = - sin (2 t) 2 t "= - 2 sin (2 t) ⇒ φ" "t = - 2 sin (2 t)" = - 2 sin (2 t) "= - 2 cos (2 t) · (2 ​​t)" = - 4 cos (2 t) ψ "(t) = (t 2)" = 2 t ⇒ ψ "" (t) = ( 2 t) "= 2

Hence we get that

y "" x = ψ "" (t) φ "(t) - ψ" (t) φ "" (t) φ "(t) 3 = 2 · - 2 sin (2 t) - 2 t · (- 4 cos (2 t)) - 2 sin 2 t 3 = = sin (2 t) - 2 t cos (2 t) 2 sin 3 (2 t)

Answer: y "" x = sin (2 t) - 2 t · cos (2 t) 2 s i n 3 (2 t)

Finding derivatives of higher orders with parametrically specified functions is carried out in a similar way.

If you notice an error in the text, please select it and press Ctrl + Enter

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

Let there be given two functions of the variable

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

Suppose 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 exclusion. When considering functions defined parametrically, the exclusion of a parameter is not only not necessary, but also not always practically possible.

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

Let's look at some examples.

Example 1. Let be an arbitrary point of a circle centered at the origin and radius R. Cartesian coordinates x and y of this point are expressed in terms of its polar radius and polar angle, which we denote here by t, as follows (see Ch. I, § 3, p. 3):

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

If Eqs. (75) are squared and added term by term, then by virtue of the identity the parameter will be excluded and the equation of the circle in the Cartesian coordinate system will be obtained, which determines two elementary functions:

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

Example 2. Consider simultaneously the ellipse

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

To each point M of the ellipse, we associate a point N of a circle that has the same abscissa as a point M, and is located with it on one side of the Ox axis. The position of the point N, and therefore of the 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 the point M from the equation of the ellipse:

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

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

Here, the parameter t ranges from 0 to.

Example 3. Consider a circle centered at point a) and radius a, which obviously touches the abscissa axis at the origin (Fig. 139). Suppose this circle is rolling without sliding along the abscissa axis. Then the point M of the circle, which coincided at the initial moment with the origin, describes a line called a cycloid.

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

Due to the fact that the circle rolls along the axis without sliding, the length of the OB segment is equal to the length of the BM arc. 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 Consequently,

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

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

Parametric definition of lines is especially often used in mechanics, where time plays the role of a parameter.

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

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

These equations are constants. When t changes, the coordinates at the point of the trajectory of the projectile 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 get the equation of the trajectory of the projectile in the form This is the equation of a parabola.

Derivative of an implicit function.
Derivative of a parametrically given function

In this article, we will consider two more typical tasks that are often found in control works on higher mathematics... In order to successfully master the material, it is necessary to be able to find derivatives at least at an intermediate level. You can learn how to find derivatives from scratch in two basic lessons and Derivative of a complex function... If everything is in order with the skills of differentiation, then let's go.

Derivative of an implicit function

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

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

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 explicitly form. What does it mean? Let's arrange a debriefing using specific examples.

Consider the function

We see that on the left we have a lonely "game", and on the right - only "x"... That is, the function explicitly expressed in terms of the independent variable.

Consider another function:

Here the variables are also "mixed". And impossible in any way express "game" only through "x". What are these methods? Transfer of terms from one part to another with a sign change, putting it out of brackets, throwing multipliers according to the rule of proportion, etc. Rewrite the equality and try to express the "game" in an explicit form:. You can twist and twist the equation for hours, but you can't.

Let me introduce you: - example implicit function.

In the course of mathematical analysis, it was proved that the implicit function exists(but not always), it has a graph (just like a "normal" function). The implicit function has 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 an implicit function. It's not that hard! All rules of differentiation, table of derivatives elementary functions remain valid. The difference is in one peculiar moment, which we will consider right now.

Yes, and I'll let you know good news- the tasks discussed below are performed according to a rather rigid and clear algorithm without a stone in front of three tracks.

Example 1

1) At the first stage, we put the finishing touches on both parts:

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

3) Direct differentiation.
How to differentiate and perfectly understandable. What to do where there are "games" under the strokes?

- just outrageously, 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 "igrek". But, the fact is that there is only one letter "igrek" - ITSELF IS A FUNCTION(see definition at the beginning of the lesson). Thus, sine is an external function, an internal function. We use the rule of differentiation of a complex function :

We differentiate the product according to the usual rule :

Note that - is also a complex function, any "game with bells and whistles" is a complex function:

The design of the solution itself should look something like this:


If there are brackets, then open them:

4) On the left side, we collect the terms in which there is a "game" with a prime. V right side- we transfer everything else:

5) On the left, we take the derivative out of the parentheses:

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

Derivative found. Ready.

It is interesting to note that you can implicitly rewrite any function. For example, the function can be rewritten like this: ... And differentiate it according to the algorithm just considered. In fact, the phrases "implicit function" and "implicit function" differ in one semantic nuance. The phrase "implicitly defined function" is more general and correct, - this function is set implicitly, but here you can express the "game" and represent the function in an explicit form. The phrase "implicit function" is understood as a "classical" implicit function, when the "game" cannot be expressed.

Second solution

Attention! The second method can be found only if you know how to confidently find partial derivatives... Beginners in calculus and dummies, please do not read and skip this paragraph, otherwise the head will be a complete mess.

Let's find the derivative of the implicit function in the second way.

We transfer all the terms to the left side:

And consider a function of two variables:

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

Thus:

The second solution allows you to check. But it is undesirable to formulate them with a clean version of the task, since the partial derivatives are mastered later, and the student studying the topic "Derivative of a function of one variable" does not seem to know the partial derivatives.

Let's look at a few more examples.

Example 2

Find the derivative of an implicit function

We put the finishing touches on both parts:

We use the linearity rules:

Find derivatives:

Expanding all brackets:

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

Final answer:

Example 3

Find the derivative of an implicit function

Complete solution and a sample design at the end of the lesson.

It is not uncommon for fractions to appear 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 an implicit function

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

Differentiate using the rule of differentiation of a complex function and the rule of differentiation of the private :

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 is. Multiply on . In detail, it will look like this:

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

On the left, we put out of the parenthesis:

Final answer:

Example 5

Find the derivative of an implicit function

This is an example for a do-it-yourself solution. The only thing in it, before getting rid of the fraction, you will first need to get rid of the three-story structure of the fraction itself. Complete solution and answer at the end of the tutorial.

Derivative of a parametrically given function

Do not strain, in this paragraph everything is also quite simple. You can write a general formula for a parametrically defined function, but in order 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 braces, but sequentially:,.

The variable is called a parameter and can take values ​​from "minus infinity" to "plus infinity". Consider, for example, a value and substitute it into both equations: ... Or humanly: "if x is equal to four, then y is equal to one." A point can be marked 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 "te" parameter. As for the "ordinary" function, for the American Indians of a parametrically defined function, all rights are also respected: you can plot a graph, find derivatives, etc. By the way, if there is a need to plot a graph of a parametrically given 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 this does not matter, because to find the derivative of a parametric function, there is a formula:

Find the derivative of the "game with respect to the te variable":

All the rules of differentiation and the table of derivatives are, of course, also valid for the letter, thus there is no novelty in the process of finding derivatives... Just mentally replace all the "x" in the table with the letter "te".

Find the derivative of "x with respect to the te variable":

Now it only remains to substitute the found derivatives into our formula:

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

As for the designations, in the formula, instead of writing, it could simply be written without a subscript, since this is the "usual" derivative "by x". But in the literature there is always a variant, so I will not deviate from the standard.

Example 6

We use the formula

In this case:

Thus:

A 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 considered example, when I found it, I expanded the parentheses under the root (although I could not do this). Chances are great that when substituted 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 parametrically defined function

This is an example for a do-it-yourself solution.

The article The simplest common problems with a derivative we considered examples in which it was required to find the second derivative of a function. For a parametrically given function, you can also find the second derivative, and it is found by the following formula:. It is quite obvious that in order to find the second derivative, one 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 the sake of simplification, we use the trigonometric formula:

Let the function be defined parametrically:
(1)
where is some variable called a parameter. And let the functions and have derivatives at some value of the variable. Moreover, the function also has an inverse function in some neighborhood of the point. Then function (1) has a derivative at the point, which, in a parametric form, is determined by the formulas:
(2)

Here and are derivatives of functions and with respect to a variable (parameter). They are often written as follows:
;
.

Then system (2) can be written as follows:

Proof

By condition, the function has an inverse function. Let's denote it as
.
Then the original function can be represented as a complex function:
.
Let us find its derivative using the rules for differentiating complex and inverse functions:
.

The rule is proven.

Proof in the second way

Let us find the derivative in the second way, based on the definition of the derivative of the function at the point:
.
Let's introduce the notation:
.
Then the previous formula takes the form:
.

We will use the fact that the function has an inverse function in the vicinity of the point.
Let us introduce the notation:
; ;
; .
Divide the numerator and denominator of the fraction by:
.
At , . Then
.

The rule is proven.

Higher order derivatives

To find derivatives of higher orders, it is necessary to carry out differentiation several times. Suppose we need to find the second-order derivative of a parametrically defined function of the following form:
(1)

Using formula (2), we find the first derivative, which is also determined parametrically:
(2)

Let's denote the first derivative by means of a variable:
.
Then, to find the second derivative of a function with respect to a variable, you need to find the first derivative of a function with respect to a variable. The dependence of a variable on a variable is also defined parametrically:
(3)
Comparing (3) with formulas (1) and (2), we find:

Now let's express the result in terms of functions and. To do this, we substitute and apply the formula for the derivative of the fraction:
.
Then
.

From this we get the second derivative of the function with respect to the variable:

It is also parametric. Note that the first line can also be written like this:
.

Continuing the process, you can get the derivatives of the function of the variable of the third and higher orders.

Note that it is possible to omit the notation for the derivative. It can be written like this:
;
.

Example 1

Find the derivative of a function defined parametrically:

Solution

Find derivatives with respect to.
From the table of derivatives we find:
;
.
We apply:

.
Here .

.
Here .

The desired derivative:
.

Answer

Example 2

Find the derivative of the function expressed in terms of a parameter:

Solution

Let's expand the brackets using formulas for power functions and roots:
.

Find the derivative:

.

Find the derivative. To do this, we introduce a variable and apply the formula for the derivative of a complex function.

.

Find the desired derivative:
.

Answer

Example 3

Find the derivatives of the second and third orders of the function given parametrically in example 1:

Solution

In Example 1, we found the first-order derivative:

Let us introduce the notation. Then the function is derivative with respect to. It is specified parametrically:

To find the second derivative with respect to, we need to find the first derivative with respect to.

Differentiate by.
.
We found the derivative with respect to example 1:
.
The second-order derivative with respect to is equal to the first-order derivative with respect to:
.

So, we have found the second-order derivative in parametric form:

Now we find the third-order derivative. Let us introduce the notation. Then we need to find the first-order derivative of the function, which is given parametrically:

Find the derivative with respect to. To do this, rewrite in an equivalent form:
.
From
.

The third-order derivative with respect to is equal to the first-order derivative with respect to:
.

Comment

You can omit the variables and, which are derived from and, respectively. Then you can write it like this:
;
;
;
;
;
;
;
;
.

Answer

In parametric representation, the second-order derivative has the following form:

Derivative of the third order.