1. Ellipse
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THE MINISTRY OF EDUCATION AND SCIENCE
OF THE REPUBLIC OF KAZAKHSTAN
S.Seifullin Kazakh Agro Technical University
Aruova A.B.
EDUCATIONAL- METHODICAL COMPLEX
on the Mathematics I
for students of the specialty 5B071900 «Radio engineering, electronics and
telecommunications»
Аstana
2012
The course content
The name of the theme
LIST OF LECTURES
hours
Literature
1. Determinants and their
properties. Matrix. Operations
with matrices. The inverse
matrix. Systems of linear
equations
2
[1],[4]
Current
control,
points
0,4
2. The simplest problem of
analytic geometry. Equations of a
straight line on a plane.
3. Analytic geometry in space.
Vectors. Simple operations with
vectors. The scalar, vector and
mixed product of vectors.
4. The curves of the second order.
Canonical equation of second
order curves.
5. Function. Methods of doing
functions. The limit function.
Fundamental theorems on limits.
Infinitely small and infinitely
large quantities. The ends.
6. The derivative of the function.
Geometric
and
mechanical
meaning.
Table of derivatives. The
differential of a function.
Derivatives of complex functions.
7. Rolle's theorem, Lagrange,
Cauchy. L'Hopital's rule.
1
[2],[5]
0,4
1
[2],[5]
0,4
1
[1],[4]
0,4
1
[1],[4]
0,4
1
[1],[4]
0,4
1
[1],[4]
0,4
8. Investigation of the function.
Extremum of the function.
Necessary
and
sufficient
conditions for the existence of an
extremum. Convexity, concavity
and inflection points. Assimptoty.
The overall study of design
features.
1
[1],[4]
0,4
9. Primitive. Indefinite integral
and its properties. Table of
integrals. Direct integration,
integration with the change of
variables, and by parts.
10. Integration of simple rational
fractions. Integration of rational
fractions.
11. Integration of expressions
containing
trigonometric
functions. Integration of irrational
functions.
12. The definite integral.
Problems leading to the definite
integral. The Newton-Leibniz.
1
[1],[4]
0,4
1
[1],[4]
0,4
1
[1],[4]
0,4
1
[1],[4]
0,4
13.
Applications
to
the
computation of the integrals of
plane figures areas. Calculation
the arc length, the amount of
body rotation. The improper
integral.
14. Complex numbers. Complex
numbers in trigonometric and
exponential form.
1
[1],[4]
0,4
1
[1],[4]
0,4
Total
15
The name of the theme
6
LIST OF PRACTICAL TASKS
hours
Literature
1. Determinants and their
properties. Matrix. The inverse
matrix. Systems of linear
equations.
2. The simplest problem of
analytic geometry. Equations of a
straight line on a plane.
2
[1], [3], [6]
Current
control,
points
0,4
2
[2], [4], [5]
0,4
3. Vectors. Simple operations
with vectors. The scalar, vector
and mixed product of vectors.
2
[2], [4], [5]
0,4
4. The curves of the second order.
The circle, ellipse, parabola.
2
[3], [4], [5]
0,4
5. Function. Methods of doing
functions. The limit function.
Fundamental theorems on limits.
Infinitely small and infinitely
large quantities. The ends.
6. The derivative of the function.
Geometric
and
mechanical
meaning. Table of derivatives.
The differential of a function.
Derivatives of complex functions.
2
[1], [3], [4]
0,4
2
[1], [3], [4]
0,4
7. Rolle's theorem, Lagrange,
Cauchy. L'Hopital's rule.
2
[1], [3], [4]
0,4
8. Investigation of the function.
Extremum of the function.
Necessary
and
sufficient
conditions for the existence of an
extremum. Convexity, concavity
and inflection points. Assimptoty.
9. Primitive. Indefinite integral
and its properties. Table of
integrals. Direct integration,
integration with the change of
variables, and by parts.
10. Integration of simple rational
fractions. Integration of rational
fractions.
11. Integration of expressions
containing
trigonometric
functions. Integration of irrational
functions.
12. The definite integral.
Problems leading to the definite
integral. The Newton-Leibniz.
2
[1], [3], [4]
0,4
2
[1], [3], [4]
0,4
2
[1], [3], [4]
0,4
2
[1], [3], [4]
0,4
2
[1], [3], [4]
0,4
13. Calculation the area of plane
figures using the definite integral.
2
[1], [3], [4]
0,4
14. Calculation the arc length, the
amount of body rotation. The
improper integral.
2
[1], [3], [4]
0,4
15. Complex numbers. Complex
numbers in trigonometric and
exponential form.
2
[1], [3], [4]
0,4
Total
30
6
LECTURE 1-2
DETERMINANTS AND THEIR PROPERTIES. MATRIX. OPERATIONS
WITH MATRICES. THE INVERSE MATRIX. SYSTEMS OF LINEAR
EQUATIONS
LECTURE PLAN:
1.
2.
3.
4.
Determinants and their properties
Мatrices
Nonsingular matrices
Systems of linear equations
Determinants and their properties
Determinants of second and third order
Consider the system of two linear algebraic equations in two variables
a11 x1 a12 x2 b1 ,
a21 x1 a22 x2 b2 ,
(1)
where a11,a12,a21, and a22 are coefficients, b1 and b2 are right-hand sides, and x1 and
x2 are unknowns.
Let us solve this system by the school method of algebraic addition, namely,
multiply the first and second equations by а22 and – a12 , respectively, and sum the
results. The coefficient of х2 will vanish. The remaining unknown, х1, is found as
follows:
x1(a11a22–a12a21)=b1a22–a12b2,
x1
b1 a 22 a12 b2
a11 a 22 a12 a 21
In a similar manner, multiplying the first equation of the system by – а21, the
second by а11 , and summing the resulting equations term by term, we obtain
x 2 (a11 a 22 a12 a 21 ) b2 a11 a 21 b1 ,
x2
b2 a11 a 21b1
a11 a 22 a12 a 21
Definition. The number
a11
a12
a21 a22
a11a22 a12a21
is called a determinant of second order.
The numbers a11,a12,a21 and a22 are called the elements of the determinant.
The second-order determinant is equal to the product of the elements of the main
diagonal minus the product of the elements of the secondary diagonal.
Example . Find the determinants
1 2
1 5 3 2 1
3 5
The unknowns х1 and х2 of the linear system (1) are determined by the formulas
b1
x1
b2 a 22
b1 a 22 a12 b2
x
1
a11 a12
a11 a 22 a12 a 21
a 21 a 22
a11
x2
a12
b1
a
b2
b2 a11 a21b1
x
21
2
a11a22 a12 a21 a11 a21
a21 a22
The determinant is called the principal determinant of the system; it is
formed by the coefficients of the unknowns. x1 and x2 are auxiliary
determinants; they are obtained by replacing the elements of the first and second
columns by the free terms of system (1).
Example.
2 3
2 x1 3x2 4,
8 1(3) 11,
1
4
x
4
x
9
.
1
2
x1
x 2
4 3
9
4
2 4
16 27 11 ,
18 4 22 ,
1 9
A third-order determinant is the number
x1 11
1,
11
x
22
x2 2
2.
11
x1
a11 a12
a13
a21 a22 a23 a11 a22 a33 a12 a23 a31 a13 a21 a32
a31 a32 a33
a13 a22 a31 a12 a21 a33 a11a23 a32 .
The simplest method for calculating a third-order determinant is the triangle rule.
The main diagonal of the determinant is the line containing the elements
a11,a22 and a33.
The secondary diagonal is the line containing the elements a13,a22 and a31.
The products of the main diagonal elements and of the elements contained in
the triangles shown below are summed with the plus sign:
a11 a12 a13
a21 a22 a23
a31 a32 a33
The products of the secondary diagonal elements and of the elements contained in
the triangle shown below are summed with the minus sign:
a11
a12
a21 a22
a31 a32
a13
a23 .
a33
Another method for calculating a third-order determinant is as follows. We
write the first columns on the right of the determinant, and sum the products of the
elements of the main diagonal and of the two parallel diagonals with the plus sign.
Then we add the products of the elements of the secondary diagonal and of the two
parallel diagonals with the minus sign
+ + +
а11
а12
а13 а11
а12
а 21
а 22
а 23 а 21
а 22
а 31
а 32
а 33 а 31
а 32
–
–
–
As a result, we obtain the determinant.
Example. Calculate the determinant by the triangle rule:
1 2
2
3
1
4 3 8 6 2 4 18 5
3
.
1
1
The determinant of order n is the expression
a11
a12
a13
...
a1n
a21
a22
a23
... a2 n
a31
a32
a33
... a3n
...
...
...
...
an1
an 2
an3
...
... ann .
An nth-order determinant contains n2 elements.
The subscript, i, indicates the number of the row and the second subscript j,
indicates the number of the column containing the element аij.
Properties of determinants. All determinants of any order have the same
properties. For simplicity, we give only properties of third-order determinants.
1. The interchange of rows and columns in a determinant does not change its
value:
a11
a12
a21 a22
a31 a32
a13
a11
a21
a31
a23 = a12
a33 a13
a22
a23
a32 (the transposed determinant).
a33
2. The interchange of two rows (columns) in a determinant changes only the
sign of the determinant:
a11
a12
a13
a12
a11
a13
a21
a22
a23 a22
a21
a23
a31
a32
a33
a31
a33
a32
.
3. If all elements of any lines (a row or a column) are zero, then the
determinant is equal to zero:
a11
a12
a13
a21
a22
a23 0.
0
0
0
The proof follows from the triangle rule.
4. A determinant containing two equal lines is equal to zero:
a11
a12
a11
a 21
a 22
a 31
a 32
a 21 =0.
a 31
5. The common multiplier of all elements of a line can be factored out:
a11
a12
a13
a11
a12
a13
ka 21
a 31
ka 22
a 32
ka 23 k a 21
a 33
a 31
a 22
a 32
a 23 .
a 33
6. A determinant containing two proportional lines, is equal to zero:
a11
a21
a31
a11 a13
a11 a11 a13
a21 a23 a21 a21 a23 0.
a31 a33
a31 a31 a33
7. If each element of some line is the sum of two terms, then such a
determinant equals the sum of two determinants, which contain these terms instead
of the elements of the lines.
a11
a12
b1 c1
a11
a12
b1
a11
a12
c1
a21
a22
b2 c2 a21
a22
b2 a21
a22
a31
a32
b3 c3
a32
b3
a32
c2 .
c3
a31
a31
8. A determinant does not change under the replacement of any line by the
sum of this line and any parallel line multiplied by some number.
Algebraic complements and minors.
Definition. The minor of an element аij is the determinant of order lower by
one consisting of the elements that remain after the deletion of the ith-series and
jth-column, which intersect in aij.
For example, the minor of the element a32 is
a11
a12
a13
a
a13
a 21 a 22 a 23 ; 11
M 32 ;
a 21 a 23
a 31 a 32 a 33
a12 a13
M 21 is the minor of а 21 .
a32 a33
Definition. The algebraic complement of an element аij is the minor of aij
multiplied by -1 raised to the proper equal to the sum of the numbers of the row
and the column intersecting in the given element:
Aij (1)i j M ij .
9. A determinant equals the sum of products of all elements of any lines and
the corresponding algebraic complements.
a11
a12
a13
a 21
a 31
a 22
a 32
a 23 a 21 A21 a 22 A22 a 23 A23 .
a 33
For a kth order determinant, we can write property 9 in the form of expansion
along the kth-column:
а11 а12 ... а1k ... а1n
a 21 a 22 ... a 2 k ... a 2n
n
a 31 a 32 ... a 3k ... a 3n a ik Aik a1k A1k a 2 k A2 k ... a nk Ank .
...
...
a n1
a n2
...
...
...
... a nk
i 1
...
... a nn
10. The sum of the products of the elements of any line and the algebraic
complements of the corresponding elements of a parallel line equals zero:
a11
a12
...
a21
a22
... a2 k
... a2 m
... a2 n
a31
a32
... a3k
... a3m
... a3n
...
...
...
...
...
an1
a1k
...
...
a1m
...
...
a1n
...
an 2 ... ank ... anm ... ann
= a1k A1m a 2 k A2 m ... a nk Anm 0
Examples. Let us expand the following determinant along the third row:
1 2 2
3
1 4 1(1)13
1
1 3
2 2
1 4
1(1) 23
1 2
3
4
3(1) 33
1 2
3
1
6 (10) 3(7) 5.
Systems of Linear Equations
Consider system of m linear equations with n unknowns:
a11 x1 a12 x2 ... a1n xn b1 ,
a x a x ... a x b ,
21 1 22 2
2n n
2
(2)
..............................................
am1 x1 am 2 x2 ... amn xn bm .
Definition. The numbers 1 , 2 ,..., n are called a solution of system
(2) if substituting them into the equations, we obtain true equalities.
Definition. A system of equations (2) is said to be consistent if it has at least
one solution, and it is said to be inconsistent if it has no solutions.
Definition. A system is called determined if it has a unique solution, and it
is called undetermined if it has many solutions.
For example, the system of equations
2 x1 3x2 5,
has no solutions, i.e., it is inconsistent, because its left2
x
3
x
6
,
1
2
hand sides are equal, while the right-hand sides are different.
x1 x2 2,
is consistent, but undetermined, because it has infinitely
3x1 3x2 6,
many solutions. Reducing the second equation by 3, we obtain two identical
equations.
Consider the following system of n linear equations with n unknowns
a11 x1 a12 x2 ... a1n xn b1 ,
a x a x ... a x b ,
21 1 22 2
2n n
2
(3)
..........
..........
..........
..........
......
an1 x1 an 2 x2 ... ann xn bn .
It is required to find a solution х1 , х2 ,..., хn of system (3), expressed in terms of
the coefficients
a ij
and the free terms
bi , where i, j 1, n
(from 1 to n).
Cramer’s Rule
To solve system (3), we multiply the first equation by А11, the second by А21,
etc., the last equation is multiplied by Аn1. Then, we sum the equations and collect
similar terms:
(a11 A11 a 21 A21 ... a n1 An1 ) x1 (a12 A11 a 22 A21 ... a n 2 An1 ) x 2 ...
(a1n A11 a 2n A21 ... a nn An1 ) x n b1 A11 b2 A21 ... bn An1
Consider the nth-order determinant composed of the coefficients of system
(3):
a11
a12
a13
... a1n
a 21
a 22
a 23
... a 2 n
a 31
a 32
a 33
...
...
...
... a 3n .
... ...
a n1
a n2
a n3
... a nn
(4)
The coefficient of х1 is the sum of the products of the element of the first column
and their algebraic complements. According to property 9, it equals determinant
(4).
The coefficients of the unknowns х 2 , х3 ,..., х n are the products of the
elements of the second, third, …, nth columns by the algebraic complements of the
elements of the first column; consequently, they equal zero by property 10.
The right-hand side is the product of the free terms and the algebraic
complements of the elements of the first column; consequently, it equals the
determinant (4) in which the first column is replaced by the column of free terms:
х1
х1 х1 ,
х1
.
Expressions for the other unknowns are obtained in a similar way: we
multiply system (3) by the algebraic complements of the n corresponding columns
хi
хi
, i 1, n ,
(5)
where is the principal determinant of the system and the хi are the auxiliary
determinants obtained from the principal one by substituting the free term column
for the ith columns.
Example. Solve the system of equations
2 x1 3 х2 х3 2,
x1 5 х2 4 х3 5,
4 x х 3 х 4.
3
1 2
Let us evaluate the principal determinant of the system:
2 3 1
0
0
1
9 7
1 5 4 9 7 4 1(1) 4
72 70 2 .
10 8
4 1 3 10 8 3
To obtain zeros in the first row, we leave the third column unchanged;
multiply it by –2 and add to the first column; then multiply it by 3 and add to the
second column.
Let us calculate the auxiliary determinants. х1 is derived from
by
replacing the first column by the free terms:
2
3
1
0
0
1
x1 5
5
4 3 7 4 24 14 10.
4
1
3
х2 is derived from
2 8 3
by replacing the second column by the free terms:
2
2
1
0
0
х2 1 5 4 9
4 4 3
1
3 4 18 30 12.
10 2 3
х3 is derived from by replacing the third column by the free terms. Zeroes are
obtained in the third row by adding the second column multiplied by –4 and 4 to
the first and third columns, respectively:
2 3
2
14
3 10
х3 1
5
5 19
5
15 (210 190) 20.
4
1
4
1
0
0
By Cramer's rule (5) we obtain
х1
х
20
х1 10
х
12
10.
5 ; х2 2
6 ; х3 3
2
2
2
1. In (5), the principal determinant must be different from zero. In this case,
system (3) has a unique solution.
2. If =0 and one of the auxiliary determinants does not equal zero ( x
0), then the system has no solutions at school, we would say that division by zero
is not allowed.
3. If =0 and all of the auxiliary determinants equal zero ( xi=0), then the
system has infinitely many solutions.
Matrices and Operations on Them
Definition. A matrix of size m n is a table written in parentheses or
between two pairs of parallel lines:
a11
a
А 21
...
a m1
a12
a1n
... a 2 n
... ...
... a mn
...
a 22
...
a m2
a11
a12
...
a 21
a 22
... a 2 n
...
...
a m1
a m2
...
a1n
...
... a mn
,
m is the number of rows, n is the number of columns, and аij is an element of the
matrix.
If m=n, then the matrix is square.
A null-matrix is a matrix А, in which all elements are zero:
0
0
О 0 0 ; О .
...
0 0
0
Definition. A square matrix А is said to be diagonal if its main diagonal may
contain nonzero elements, but all nondiagonal elements are zero:
a11
0
В
...
0
0
0
...
0
... ...
... a mт
.
...
a 22
...
0
Definition. The identity matrix is the diagonal matrix with ones on the
diagonal:
1 0 ... 0
Е
0
1 ... 0
... ... ... ...
0
0 ... 1 .
Definition. Matrices of the same size are equal if their respective elements
are equal.
Two matrices of the same size can be added (elementwise):
a11 a12
a
a
А B 21 22
... ...
am1 am 2
... a1n b11 b12
... a2 n b21 b22
... ... ... ...
... amn bm1 bm 2
... b1n
c11 c12
... b2n
c21 c22
C
... ...
... ...
... bmn
cm1 cm 2
cij aij bij .
Any matrix can be multiplied by any number
should be multiplied by this number):
а11
а
А 21
...
а m1
а12
а 22
...
...
...
а m 2
... c1n
... c2n
,
... ...
... cmn
all elements of the matrix
а1n
... а 2n
...
... а mn
.
Subtraction is defined as
А–В=А+(–1)В.
Two matrices can be multiplied only if the number of columns in the first matrix
equals the number of rows in the second matrix. An m n matrix is multiplied by
an n k matrix as follows:
a11 a12
a21 a22
А B
... ...
am1 am 2
... a1n b11 b12
... a2 n b21 b22
... ... ... ...
... amn bn1 bn 2
...
...
...
...
n
na b
ab
b1k i 1 1i i1 i 1 1i i 2
n
n
b2 k a b a b
i 1 2i i1 i 1 2i i 2
...
...
...
n
n
bnk
ami bi1 ami bi 2
i 1
i 1
n
... a1i bik
i 1
n
... a2i bik ,
i 1
...
...
n
... ami bik
i 1
We obtain a matrix of size m k .
Multiplication of matrices is not commutative, and sometimes even impossible:
АВ ВА.
The Gauss-Jordan Method
Consider the Gauss method in the case where the number of equations
coincides with that of unknowns:
a11 x1 a12 x2 ... a1n xn b1 ,
a x a x ... a x b ,
21 1 22 2
2n n
2
..............................................
an1 x1 an 2 x2 ... ann xn bn .
(6)
Suppose that а11 0; let us divide the first equation by this coefficient:
a11 x1 a12 x 2 ... a1n x n b1
1
x1 a12 x 2 ... a1n x n b1 . (*)
а11
Multiplying the resulting equation by –а21 and adding it to the second equation of
system (6), we obtain
a 22 x 2 a 23 x3 ... a 2 n x n b2 .
Similarly, multiplying equation (*) by –аn1 and adding it to the last equation of
system (6), we obtain
a n 2 x 2 a n3 x3 ... a nn x n bn .
At the end, we obtain the new system of equations with n–1 unknowns:
a x a x ... a x b ,
23 3
2n n
2
22 2
a x a x ... a x b ,
33 3
3n n
3
32 2
................................................
an 2 x2 an3 x3 ... ann xn bn .
(7)
System (7) is obtained from system (6) by applying linear transformations of
equations; hence this system is equivalent to (6), i.e., any solution of system (7) is
a solution of the initial system of equations.
To get rid of х2 in the third, the forth, …, nth-equation, we multiply the second
equation of system (7) by
1
and, multiplying this equation by the negative
а22
coefficients of х2 and summing them, obtain
x a x а х ... a x b ,
12 2
13 3
1n n
1
1
( 2)
( 2)
( 2)
x2 а23 х3 ... a2 n xn b2 ,
( 2)
( 2)
( 2)
а33 x3 ... a3n xn b3 ,
.........................................
an 3( 2 ) x3 ... ann ( 2 ) xn bn ( 2 ) .
Performing this procedure n times, we reduce the system of equations to the
diagonal form
x a x а х ... a x b ,
12 2
13 3
1n n
1
1
x2 а23( 2 ) х3 ... a2 n ( 2 ) xn b2 ( 2 ) ,
( 3)
( 3)
x3 ... a3n xn b3 ,
.......... .......... .......... ....
ann ( n ) xn bn ( n ) .
We determine хn from the last equation, substitute it in the preceding equation and
obtain xn-1, and so on; going up, we determine х1 from the first equation. This is the
classical Gauss method.
Consider the system of m equations with n unknowns
a11 x1 a12 x2 ... a1n xn b1 ,
a x a x ... a x b ,
21 1
22 2
2n n
2
..............................................
aт1 x1 aт 2 x2 ... aтn xn bт .
(8)
Definition. The matrix composed of the coefficients of system (8) is called
the principal matrix of this system:
a11
a
А 21
...
a m1
a12
a 22
...
a m2
... a1n
... a 2 n
... ... .
... a mn
Adding the column of free terms of system (8) to this matrix, we obtain the
augmented matrix
a11
a21
А
...
a
m1
a12
a22
...
am 2
a1n b1
... a2 n b2
... ... ... .
... amn bm
...
The following linear operations on the rows of such a matrix are allowed:
- permutation of rows;
- multiplication of a row by some number and adding it to another row;
- permutation of columns (but we must remember to which unknowns they
correspond);
- no operations on columns are allowed (columns cannot be multiplied by
numbers, summed, etc).
The Gauss-Jordan method consists in reducing (by linear operation on rows)
the principal matrix to the identity matrix, i.e., to the form
1
0
...
0
0 ... 0 с1
1 ... 0 с 2
... ... ... ... .
0 ... 1 с т
If the columns were not interchanged, the solution of the system of linear
equations is
х1 с1; х2 с2 ;...; хт ст .
Examples. Solve the following system of equations by the Gauss-Jordan
method:
x1 x2 х3 1,
2 x1 x2 2 х3 1,
x x 3 х 2.
3
1 2
We compose the augmented matrix of the system and, applying linear
combinations of rows, reduce the principal matrix to the identity:
1
1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 0 1 0 0
2
2
1
2
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
,
1 1 3 2 0 0 2 1 0 0 1 1 0 0 1 1 0 0 1 1
2
2
2
1
1
х1 ; х2 1; х3 .
2
2
Rank of a Matrix
Definition. Consider two systems
1 , 2 , 3 ,..., n ;
1 , 2 , 3 ,..., n .
These systems are said to be linearly dependent if
k11 k 2 1 0,
k1 k k 2 k 0,
where k1 0 or k2 0.
If these relations hold only for k1=0 and k2=0, then these two systems are linearly
independent.
Consider an m n matrix
А
a11
a12
... a1n
a 21
a 22
... a 2 n
...
...
a m1
a m2
...
... .
... a mn
Definition. The maximal number of linearly independent rows in the matrix
А is called the rank of this matrix.
The maximal number of linearly independent rows equals the maximal
number of linearly independent columns.
Definition. A kth-order minor of a matrix А is the determinant consisting of
the elements of arbitrarily chosen k columns and k rows.
Theorem. If all kth-order minors of a matrix are zero, then all (k+j)th-order
minors are also zero.
Theorem. The rank of a matrix equals the maximal order of a nonzero
minor.
The first method for calculating the rank of matrix (the bordering method).
(а) The method is to pass lower-order minors to higher-orders minors.
(b) Choose a nonzero minor and border it by a column and a row.
(c) If all of the bordered minors are zero, then the rank of the matrix equals
the order of the nonzero kth-order minor:
rA k .
Example . Calculate the rank of a matrix
1
2 4 3
1 2 1 4
A
0 1 1 3
4 7 4 4
0
2
.
1
5
We have r A 4 , i.e., can the rank not be larger than 4.
4 3
M2
4 6 2 0 ,
2 1
4 3
1
1 3 10
M 3 2 1 4 1 1 1 1 10 11 0 ,
1 1 3
0 1 0
2 4 3
1
2 1 3 10
2 1 10 2 1 12
1
2
1
4
1
1
1
1
M4
1 1 1 1 0 0 0,
0 1 1 3
0 0 1 0
4 3 8
4 1 12
4 7 4 4 4 3 4
8
M4
4
3
2
1
1
1
7
4
0
1
3
10
4 2
1
1
1 3
0
1
0
0
3
4
8
9
1
3
1
4 5
3
1 10 3
1 11 0
1 1 3 1
3
8
9
0 0.
3 11 0
0
Since all of forth-order minors equal zero and the determinant of third order
does not equal zero, it follows that r A 3 .
The second method for calculating the rank of a matrix.
The elementary transformations of matrices include:
- permutation of columns or rows in a matrix,
- multiplication and division of the elements of a row or a column by a
nonzero number,
- linear combination of rows (columns).
The elementary transformations of a matrix do not affect its rank.
Theorem. The rank of a matrix equals the number of nonzero elements on
the main diagonal of any diagonal matrix obtained by elementary transformations
from the initial matrix. (Without proof).
In the Gauss-Jordan method, only operations on rows are allowed; in
calculating rank, we may operate on both rows and columns.
2 4 3 1
1 2 1 4
A
0 1 1 3
4 7 4 4
0 1 2 1 4 2
2 0 0 1 9 4
…
1 0 1 1 3 1
5 0 1 0 12 3
We interchanged the first and second rows and obtained zeroes in the first
column by using the second row. Then, using the first column, which has only one
nonzero element, we obtained zeroes in the first row. Interchanging the second and
third rows and using the new second row, we obtain zeroes in the second column,
etc.
1
0
0
0
0
0
0
0
1
9
1 1
1
0
3
12
1
0
0
0
0 1 0 0
4 0 1 1
1 0 0 1
3 0 0 1
0 0 0 0 1 0
1 0 0 0 0 1
0 1 9 4 0 0
0 0 0 0 0 0
0
3
9
9
0
0
1
0
0 1
1 0
4 0
4 0
0 0
0 0
;
0 0
0 0
0
1 0 0 0
0 1 9 4
0 1 9 4
0 0 0
rA 3.
Kronecker-Capelli theorem. Consider the system of m linear equations
with n unknowns:
a11 x1 a12 x2 a13 x3 ... a1n xn b1 ,
a x a x a x ... a x b ,
21 1 22 2
23 3
2n n
2
..........................................................
am1 x1 am 2 x2 am3 x3 ... amn xn bm .
(8)
Let us compose the principal and augmented matrices of this system:
a11
a
А 21
...
a m1
a12
a 22
...
a m2
... a1n
... a 2 n
... ... ;
... a mn
a11
a
A 21
...
a m1
a12
a 22
...
a m2
... a1n b1
... a 2 n b2
... ... ... .
... a mn bm
Theorem. A system of equations is consistent if and only if the rank of the
augmented matrix A equals that of the principal matrix А, i.e.,
rA rA .
Corollary 1. Any homogeneous system of equations
(9)
a11 x1 a12 x2 a13 x3 ... a1n xn 0,
a x a x a x ... a x 0,
21 1 22 2
23 3
2n n
.........................................................
am1 x1 am 2 x2 am3 x3 ... amn xn 0,
is consistent. Indeed, it suffices to take bi=0 for all i.
Corollary 2. System of equations (8) is consistent and has a unique solution
if and only if the ranks of the matrices are equal and coincide with the number of
unknowns.
Example. Determine whether the following system is consistent:
5 x1 x2 2 х3 х4 7,
2 x1 x2 4 х2 2 х4 1,
x 3 x 6 х 5 х 0.
2
3
4
1
1 1 3 6
5 1 2
А 2 1
4 2 0 14 32
1 3 6 5 0 7 16
1 0 0
0 1
16 12
0 1
0
7
7
0
0
0
0
0
5 1 0 0
0
24 0 7 16 12
12 0 0 0
0
0 0 0
1 0 0 ,
0 0 0
1 7 1 5
2
1
7
5 1 2
А 2 1
4 2 1 0
7
6 1 8
1 3 6 5 0 0 14 12 2 21
0 1 0 0 0 0 1 0 0 0
1 0 0 0
0 7 6 1 8 0 1 0 0 0 0 1 0 0,
0 0 0 0 5 0 0 0 0 1 0 0 1 0
rA 2, rA 3. The system is inconsistent.
Inverse Matrices
Definition. A square matrix is said to be nondegenerate, if it has nonzero
determinant.
Consider the identity matrix
Е
1
0
0
0
1
0
0
0
.
1
Definition. The inverse matrix for А is the matrix А–1for which
А А 1 Е или А1 A Е .
(11)
Inverse matrices exist only for nondegenerate matrices.
The first method for finding the inverse matrix. Consider a square
nondegenerate matrix of size n n
a11
a
A 21
...
a n1
a12
a 22
...
a n2
... a1n
... a 2 n
... ... .
... a nn
Let us compose the square matrix
А11
А
A* 12
...
А1n
А21
А22
...
А2 n
Аn1
... Аn 2
... ... ,
... Аnn
...
where the Аij are the algebraic complements of аij.
The matrix А* is called the adjoint matrix.
Let us multiply А by А* :
na A
1i 1i
i n1
a A
2i 1i
A A* i
1
...
n
ani A1i
i 1
n
a1i A2i
i 1
n
a2i A2i
i 1
...
n
ani A2i
i 1
... a1i Ani
| A | 0 0
i 1
n
... a2i Ani 0 | A | 0 .
i 1
...
...
n
0
0 | A |
... ani Ani
i 1
n
The remaining elements of the first row in the new matrix are the sum of the
products of the elements of the first row and the algebraic complements of the
parallel rows. According to property 10, they equal zero. The same is true for the
elements of the second row, except the second element, which are the sum of the
products of the elements of the second row by the algebraic complements of the
first, third , ... rows.
To obtain the identity matrix from this diagonal matrix, we must divide it by
the determinant | A | d , i.e.,
A*
A
E;
d
consequently,
A
A 1
A*
,
d
1
А11
d
А12
d
...
А1n
d
А21
d
А22
d
...
А2 n
d
AA 1 E ,
...
...
...
...
Аn1
d
Аn 2
d
...
Аnn
d
.
(12)
Example. Find the inverse matrix for the matrix
1 2 1
А 0 1 1 ,
2 0
1
A11 (1)11
A13 (1)13
1 1
0
2
A31 (1) 31
1
1 ,
2,
0
1 1
1
2
1
1
1
2
0 1
1
1 3 4 1 ,
3
A12 (1)1 2
A21 (1) 21
0 1
2 1
2 1
0
1 2 3 , A23 (1) 23
2
1
33
2
d | A | 0 1
0 4
0 1
A22 (1) 2 2
A33 (1)
1
1,
1 ,
A32 (1) 3 2
А 1
1
2,
2 ,
1 2
2 0
1 1
0
1
4,
1 ,
1 2 1
2
3 1 .
2
4 1
Let us verify condition (11) from the definition of the inverse matrix:
1 2 1 1 2 1 1 4 2 2 6 4 1 2 1 1 0 0
АA 1 0 1 1 2 3 1 2 2
3 4
1 1 0 1 0 .
2 0 1 2 4 1 2 2
44
2 1 0 0 1
The second method for finding the inverse matrix. This method is similar
to the Gauss-Jordan method. Applying the Jordan transformations to the rows of
the principal matrix, we reduce it to the identity matrix, and the identity matrix
added to it becomes the inverse matrix.
а11
а 21
...
а
n1
а12
а 22
...
... а1n 1
... а 2 n 0
...
а n2
0 ...
1 ...
... ... ... ...
... а nn 0 0 ...
0
0
...
1
A11 A12
A21 A22
... ...
An1 An 2
... A1n
... A2 n
... ...
... Ann
Example.
1 2 1 1 0 0 1 2 1 1 0 0 1 0
0
1
1
0
1
0
0
1
1
0
1
0
0 1
2 0 1 0 0 1 0 4 3 2 0 1 0 0
1 0 1 1 2 1 1 0 0 1
0 1 1 2 3 1 0 1 0 2
0 0 1 2 4 1 0 0 1 2
0
1 0 1 0
1 2 4 1
1 1
2
3
4
2
1
1
1
.
Matrix Method for Solving Systems of Equations
Consider a system of n equations with n unknowns:
a11 x1 a12 x2 a13 x3 ... a1n xn b1 ,
a x a x a x ... a x b ,
21 1 22 2
23 3
2n n
2
..........................................................
an1 x1 an 2 x2 an3 x3 ... ann xn bn .
(13)
Let us find a solution of system (13), by using matrices.
The matrix method applies only where the number of equations equals that
of unknowns. Let us write system (13) in matrix form; for this purpose we
introduce, principal matrix А, the column matrix Х, and the column matrix of free
terms В:
a11
a21
A
...
an1
a12
a22
...
an 2
x1
... a1n
... a2 n
x2
X
;
;
...
... ...
... ann
xn
b1
b
B 2 .
...
bn
Then system (13) can be written in the form of the matrix equation
АХ=В.
Two matrices of the same size are equal if and only if each element of one
matrix equals the corresponding element of the other matrix. To find the matrix Х,
we multiply both sides of the matrix equation by the inverse matrix А-1 on the left
A 1 A X A 1 B .
Since А 1 А
Е is the identity matrix, we have
X A 1 B .
(14)
Thus, to solve the given system of equations by the matrix method, it is
sufficient to find the inverse matrix А-1 and multiply it by В on the right.
Example. Solve the following system of equations by the matrix method:
x1 x2 2 х3 1,
2 x1 x2 2 х3 4,
4 x x 4 х 2 .
3
1 2
Let us write the system in matrix form:
1 1 2 x1 1
2
1
2
x 2 4 .
4 1 4 x 2
3
Applying (14), we obtain
x1
1
1
x
A
4
2
x
2
3
.
The principal matrix is nondegenerate, and its determinant equals d=6.
Now, let us find the inverse matrix:
A 1
1
1
3
2
0
3
1
1
2
2
3
1
.
3
1
2
The solution of the system is
1
1
3
x1
2
x2 0
3
x
1
3
1
2
2
4
1
3 1
3
1
8
4 0
3
3
1 2 1 2
2
4
3 1
2
2 .
3
1 2
Thus, the solution of the system is x1= –1, x2= –2, x3=2.
LECTURE 3.
THE SIMPLEST PROBLEM OF ANALYTIC GEOMETRY. EQUATIONS
OF A STRAIGHT LINE ON A PLANE.
LECTURE PLAN:
1. The simplest problem of analytic geometry
2. Equations of a straight line on a plane
Analytic Geometry in the Plane
Consider the Cartesian rectangular coordinate system in the plane. Taking
the projection of any point М1 on the x and y coordinate we obtain two numbers
x=a1 and y=b1. Take two numbers a2 and b2 plot a2 on the x-axis and b2 on the yaxis . Having drawn two straight lines parallel to the coordinate axes through these
points we find obtain a point M2 in their intersection.
a2 M2(a2;b2)
M1(a1;b1)
b1
b2
0
a1
x
M 1 a1 , b1
Thus, there is a one–to–one correspondence between
points in the plane and pairs of numbers.
a 2 , b2 M 2
The distance between two points. Let us find the distance between two
points М1 and М2 in the plane.
y
M2(x2;,y2)
d
M1(x1;y1)
0
x
Compose the vector М1М 2 x2 x1; y2 y1 .
The length of this vector is defined by
d М 1 М 2 ( x 2 x1 ) 2 ( y 2 y1 ) 2 .
This is the distance between the two given points.
Example. Find the distance between the two points А(2;3) and
Using the above formula, we obtain
В(–4;11).
d (2 4) 2 (3 11) 2 36 64 10 .
Division of an interval in a given ratio. Suppose given an interval М1М2. Let
us find the coordinates a of point М on the interval for which
ММ 1
М 2М
.
Compose the vectors M 2 М x x 2 ; y y 2 and МM 1 x1 x; y1 y.
y
M2(x2,y2)
M(x,y)
M1(x1,y1)
0
x
This gives the x coordinate; y is found in a similar way:
x
x1 x 2
;
1
y
y1 y 2
.
1
To obtain a formula for the midpoint of the interval, we take =1:
x
x1 x 2
;
2
y
y1 y 2
.
2
Example. Given the two points М1(–2;4) and М2(6;2), find the midpoint of
the interval М1, М2.
x
М1
М (2;3)
26 4
2,
2
2
y
М2
42 6
3.
2
2
Lines and Their Equations
Definition. A line is the locus of points satisfying a characteristic condition
of this line.
y
M1 ( x1; y1 )
M n ( xn ; yn )
M 2 ( x2 ; y2 )
0
x
Definition. An equation of a line is a relation of the form
F ( x; y ) 0,
which holds for the coordinates of all points of the line, i.e.,
F( x1 , y1 ) 0 ,
F( x 2 , y 2 ) 0 ,
F ( xn , y n ) 0 ,
..............
Straight Lines in the Plane
The equation of a straight line with a slope. Given a straight line, we
denote the angle between this line and the x–axis by and the interval cut out by
the line on the x-axis by b.
у М(х;у)
y-b
B
b
0
N
x
Definition. The slope tangent of the angle between a straight line and the x–
axis is called the slope of the line and denoted by
k=tan.
Suppose that k is the slope of a line and b is its y–intercept.
The equation of a straight line with a slope has the form
y kx b .
The equation of a straight line with given slope passing through a given
point. Suppose that a straight line passes through a point М0(x0,y0) and has slope
k.
y
M(x;y)
φ
M0(x0;y0) N
φ
0
x
By analogy with the equation of a straight line with a slope consider the triangle
MN
tan for any point М on the under
М0MN; we have
M0N
consideration or
y y0
k.
x x0
Thus, the required equation is
y – y0 = k(x–x0).
The equation of a straight line passing through two points. Suppose that
a straight line passes through two points М1(х1;у1) and М2(х2;у2).
y
M(x;y)
M2(x2;y2)
M1(x1;y1)
0
x
Take a point M(x,y) on the line and consider the vectors
M 1 M x x1 ; y y1 and M 1 M 2 x 2 x1 ; y 2 y1 .
These two vectors M 1 M и M 1 M 2 lie on the same straight line and are
collinear.
The collinearity condition is the proportionality of the perspective coordinates, i.e.,
x x1
y y1
.
x2 x1 y2 y1
(10)
This is the equation of a straight line passing through the two given points.
Example. Write an equation of the straight line passing through the points
М1(2;–5) and М2(3;2) and find k and b.
Using formula (10), we obtain
x2 y5
32 25
7x–14=y+5.
Thus the equation of the straight line is
y=7x–19,
and the slope and the y–intercept are
k=7, b= –19.
The general equation of a straight line and its analysis.
Definition. A first–order equation in variables x and y determines a straight
line in the plane.
The general equation has the form
Ax By C 0 ,
where А and В are called the coefficients of the variables.
1. If the free term is С=0, then the equation has the form
Ax By 0 .
Since х=0 and y=0 satisfy this equation, it follows that the straight line passes
through the origin.
2. If the coefficient of х is А=0, then the equation has the form
By C 0 or y
C
const , i.e., the line is parallel to the x–axis.
B
3. If the coefficient of y is B=0, then the equation has the form
Ax C 0 ,
and the line is parallel to the y–axis.
4. If А=С=0, then the line
В у=0
(or у=0)
coincides with the x–axis.
5. If В=С=0, then the line
А х=0
(or х=0)
coincides with the y–axis.
2.7. 5. The two-intercept equation of a straight line. Suppose that a straight
line intersects the coordinate axes in points M1(a;0) and М2(0;b)
у
М2(0;b)
b
M1(a;0)
0
a
x
Knowing the two points М1 and М2 through which the line passes, we can write the
equation of the line in form (10):
xа y0
, or
0а b0
x y
1.
a b
This is the two-intercept equation of the line.
Example. Reduce the equation 3x 2 y 1 0 , to the two-intercept form.
Take the variables to the left-hand side
3x 2 y 1 .
We have
y
x
1
1
1 , where a and b .
1 1
3
2
3
2
The angle between two straight lines. Parallel and perpendicular lines. Suppose
given two straight lines with slopes tan 1 k1 and tan 2 k 2 .
у
=?
1 2
1
2
0
х
It is seen from the figure that the angle between the lines equals
Using the formula for the tangent of the difference between
two angels, we obtain
tan tan( 2 1 )
=2–1.
tan 2 tan 1
.
1 tan 1 tan 2
Replacing the tangents by the slopes k1 and k2, we obtain the following formula for
the tangent of the angle between two straight lines:
tan
k2 k1
.
1 k1k2
(11)
Formula (11) gives conditions for two lines to be parallel and perpendicular.
(1) Suppose that the right lines are parallel, i.e., the angle between them is
0 ; substituting it into formula (11), we obtain
tan 0 0
k2 k1
1 k1k2 .
This fraction vanishes, if k2–k1=0.
Thus, two straight lines are parallel if and only if their slopes are equal:
k2=k1.
(2) Suppose that two straight lines are perpendicular; then the angle between
them is
2
. Substituting it in (11), we obtain
tan
2
k2 k1
1 k1k2 .
This fraction equals infinity when the denominator vanishes:
1 k1 k 2 0 .
Consequently, the condition for two straight lines to be perpendicular is
k2
1
.
k1
Example. Write equations of the straight lines passing through the point
М0(1;1) and parallel and perpendicular to the line 2 x 3 y 4 0 .
Let us write the equation of the given line in the form y=kx+b:
2
4
2
4
3 y 2 x 4 , or y x ; k1 , b .
3
3
3
3
To compose an equation of a straight line, we use the formula
y y 0 k ( x x0 ) .
(*)
Since the required line must be parallel to the required one, it follows that
2
k 2 k1 .
3
Substituting this into equation (*), we obtain
2
y 1 ( x 1) , or 2 x 3 y 1 0 .
3
The perpendicularity condition gives the slope:
1
3
k2 .
k1
2
The equation of the required line is
3
y 1 ( x 1) ; 3x 2 y 5 0 .
2
The mutual arrangement of two straight lines. Given equations of two straight
lines
A1 x B1 y C1 0
and
A2 x B2 y C2 0 .
Determine conditions on the coefficient, for these right lines to intersect, be
parallel, or coincide.
1. To determine the mutual arrangement of lines, we must analyze the
system of equations
A1 x B1 y C1 0,
A2 x B2 y C2 0.
If the lines intersect, then this system has a unique solution, and its principal
determinant is nonzero:
А1
В1
А2
В2
0;
А1 В2 В1 А2 ;
А1 В1
.
А2 В2
(12)
Thus, if the straight lines intersect, then the coefficients must not be
proportional.
2. Suppose that the straight lines are parallel, i.e., they have no common
points, and the system of equations has no solution; then the principal determinant
vanishes, and the auxiliary determinants are nonzero:
А1
В1
А2
В2
0;
х 0, у 0 ;
А1 В1
;
А2 В2
С1
В1
С2
В2
0 ; С1 B2 C 2 B1 0 ; C1 В2 C 2 B1 ;
A1 B1 C1
.
A2 B2 C 2
C1 В1
;
C 2 В2
(12΄)
This is the parallelis condition.
3. When the straight lines coincide, i.e., have many common points, the
system of equations has infinitely many solutions. In this case, the auxiliary and
principal determinants are zero:
А1
В1
А2
В2
0;
х 0, у 0 ;
A1 B1 C1
.
A2 B2 C 2
(12΄΄)
This is the condition for straight lines to coincide.
Example. 1. Concider the straight lines given by
4 x 5 y 7 0,
3x 2 y 21 0.
4
5
, which means that these lines
According to formula (12), we have
3
2
intersect.
2. Consider the straight lines given by
3x 2 y 8 0,
6 x 4 y 9 0.
3 2
8
, and the lines are parallel.
6 4
9
3. Consider the straight lines given by
According to formula (12΄), we have
4 x 5 y 7 0,
12 x 15 y 21 0.
4
5
7
According to formula (12΄΄) we have
, and the lines coincide.
12 15 21
Line Pencils
Definition. A line pencil is the set of straight line passing through a given point.
Arbitrarily varying the coefficient k in the equation y y 0 k ( x x 0 ) of a line
passing through a point М0(х0,у0), we obtain equations of all straight lines passing
through the point with coordinates х0, у0.
All these straight lines constitute the line pencil centered at М0.
A line pencil can also be specified by two equations of lines from this pencil
(which determine of these lines) the common point М0(х0, у0) by the coordinates
of its center.
Suppose given intersecting straight lines with general equations
A1 x B1 y C1 0 ,
A2 x B2 y C2 0 .
Consider the equations
q1 ( A1 x B1 y C1 ) q 2 ( A2 x B2 y C2 ) 0 ,
where q1 and q2 undetermined multipliers, not both zero.
(13)
Let us prove that, for any q1 and q2, the straight line (13) passes through the point
М0 (х0; у0), which is the intersection point of the two given lines.
The coordinates of the point М0 satisfy both equations
A1 x 0 B1 y 0 C1 0 ,
A2 x0 B2 y0 C2 0.
Consequently, the coordinates of М0 satisfy equation (13), i.e.,
q1 ( A1 x0 B1 y 0 C1 ) q 2 ( A2 x0 B2 y 0 C 2 ) 0 .
Thus, the point М0 belongs to the straight line (13) for any q 1 and q2; but for
particular values of q1 and q2, we obtain some straight line from the pencil.
In practice, it is more convenient to write an equation of a pencil with one
parameter :
( A1 x B1 y C1 ) ( A2 x B2 y C 2 ) 0 .
Example. Draw a straight line through the intersection point of the lines х+у–3=0
and 2х–у+5=0 so that it passes through the point М(3;5).
Let us write the equation of the pencil of lines according to the formula. To find
the required line, we must find the value of the parameter :
x y 3 (2 x y 5) 0.
We substitute the coordinates of the point М(3;5) into the equation of the pencil:
(2 x y 5) x y 3 ,
( x y 3) (3 5 3)
5
.
2x y 5
(6 5 5)
6
Substituting in to the equation of the pencil, we obtain the required equation:
2
5
1
5
x y 3 (2 x y 5) 0 ; x y 3 1 x y 4 0;
6
3
6
6
2
5
1
x 1 y 7 0 .
3
6
6
The Normal Equation of a Straight Line
The position of a straight line in the plane is completely determined by the
distance from the line to the origin of coordinates, i.e., the length of the
perpendicular
vector
y
OT p
from the origin to the fight line, and the unit normal
Т
n
0
0
α M
x
Let us construct an equation of a straight line from the length of the perpendicular
р and its angle with the x-axis.
To this end we take an arbitrary point М on the line and find the projection of the
0
vector OM on the unit normal vector n .
The point М belongs to the given line if and only if
рr 0 OM OT p .
n
(*)
In terms of inner product of vectors, we have
рr 0 OM OM , n 0 .
n
Considering this product, we obtain
(OM , n 0 ) p 0 .
(14)
Equation (14) is the normal equation of the line in vector form. To pass to
coordinates, note that the projections of the unit normal vector are
n 0 cos ; sin ,
where is the angle between this vector and the axis ОX. Let the projections of
the radius-vectors OM on the axes ОХ and ОУ are х and у, respectively. The
inner product formula (14) implies
x cos y sin p 0 .
(15)
This equation is called the normal equation of the straight line. It is clear from
equation (15) that the “normality” conditions are
(1) cos 2 sin 2 1 (the sum of squared coefficients of х and у equals 1);
(2) –р<0 (the free element is negative).
Example. Which of the following equations are normal?
2
2
1
3
1 3
(a) x y 7 0 , 0.61 1 , this is not a normal equation;
2
5
2 5
2
2
12
5
12 5
x
y 4 0 , 1 and –4<0, this is a normal equation;
(b)
13
13
13 13
4
3
(c) x y 2 0 ,
5
5
2
2
4 3
1 but 2>0, this is not a normal equation.
5 5
A normalizing factor. Suppose given a general equation of a straight line given:
Ax By C 0 .
Definition. The normalizing factor is the number such that the equation
multiplied by this number is normal.
Let us multiply the general equation of a straight line by number :
Ax By C 0 .
By definition, if this equation is normal, then the two normality conditions hold:
(A)2 (B)2 1 ; C 0 .
Removing parentheses, we obtain
2 ( A2 B 2 ) 1 ;
extracting the root, we see that the normalizing factor is
1
A2 B 2
;
the sign of the fraction is opposite to that of the free element of the
equation.
Example. Reduce the equation 8х–6у+5=0 to the normal form.
We find the normalizing factor
general
1
1
10
64 36
and multiply the equation by this factor term by term:
8
6
5
x y
0
10
10
10
or
4
3
1
x y 0.
5
5
2
The Distance Between a Point and a Straight Line
Suppose given a point М0(х0;у0) and a straight line with normal equation
xcos+ysin–p=0.
у
М0(х0;у0)
d
M1
0
x
Consider the distance from the point to the right line, which equals the distance
from the point to the intersection point of two perpendicular straight lines.
The distance is determined by the formula
d x0 cos y 0 sin p .
If a general equation of straight line is given, then the distance is determined by
the formula
d
Ах0 Ву0 С
.
A2 B 2
(16)
Example. Find the distance from the point М(1;–2) to the straight line 6х–
8у+9=0.
1
А2 В 2
1
36 64
1
,
10
6 16 9
31
6
8
9
3.1 .
х
у 0; d
10 10 10 10
10
10
10
LECTURE 5.
ANALYTIC GEOMETRY IN SPACE. VECTORS. SIMPLE OPERATIONS
WITH VECTORS. THE SCALAR, VECTOR AND MIXED PRODUCT OF
VECTORS
LECTURE PLAN
1. Planes
2. Vectors. Operations with vectors.
PLANES
The general equation of a plane. Suppose given a vector N { A; B; C}
perpendicular to a plane and a point M 0 ( x0 , y0 , z 0 ) in this plane. This vector is
called a normal vector.
It is required to write an equation of the plane.
N
M(x,y,z)
M0(x0,y0,z0)
According to the general scheme, we take an arbitrary point M(x,y,z) in the plane.
Consider the vector M 0 M x x0 ; y y 0 ; z z 0 . For any point in the
plane, the vector M 0 M is perpendicular to the normal vector N .
Since
M 0 M N , it follows that the scalar product vanishes:
N M 0 M 0 , or, in coordinate form,
A( x x 0 ) B( y y 0 ) C ( z z 0 ) 0 ,
Ax By Cz Ax 0 By 0 Cz 0 0 .
(*)
Denoting this numerical expression by D, we obtain
Ax By Cz D 0 .
(23)
This is the general equation of a plane; the coefficients of A,B and C of x, y, and z
are the coordinates of the normal vector N .
The three-intercept equation of a plane. Suppose given a plane not
passing through the origin but intersecting the coordinate axes at points
M 1 (a,0,0) , M 2 (0, b,0) , and M 3 (0,0, c) . Suppose also that the segments a,b,
and c are known and it is required to write the equation of the plane from the
intercepts. Let us write the general equation
Ax By Cz D 0 .
The coefficients are not known yet, so we choose them so, that the plane cuts out
the given segments a,b, and c on the coordinate axes.
Since, the point M 1 (a,0,0) belongs to the plane, its coordinates satisfy the
general equation of the plane, and
z M3
M2
M1
х
y
А а B 0 C 0 D 0.
А а D 0;
Therefore,
A
D
.
a
By analogy, we obtain
А 0 B b C 0 D 0; B b D 0 ; B
D
,
b
А 0 B 0 C с D 0; С с D 0 ;
С
D
.
с
Substituting the obtained values of the coefficients А,B, and C into the general
equation of the plane, we obtain
D
D
D
x y z D 0,
a
b
c
which gives, after the reduction by D, the three-intercept equation of the plane:
x y z
1.
a b c
The equation of a plane passing through three points. Suppose given,
three points M 1 ( x1 ; y1 ; z1 ) , M 2 ( x 2 ; y 2 ; z 2 ) , and M 3 ( x 3 ; y 3 ; z 3 ) . As is
known form elementary geometry, there exists a unique plane passing through
these points. It is required to write its equation.
Following the general scheme, of we take an arbitrary point M(x;y;z) in the
plane.
z
M2
M1
M(x;y;z)
M3
y
0
x
The characteristic feature of a plane is that if a point М belongs to the plane, then
the three vectors
М 1 М ={x–x1;y–y1; z–z1},
М 1 М 2 = {x2–x1;y2–y1;z2–z1},
М 1 М 3 ={x3–x1;y3–y1;z3–z1},
are coplanar.
Therefore, the triple product of these vectors must be zero:
М 1М М 1М 2 М 1М 3 0 .
Expressing the triple product in terns of coordinates, we obtain an equation
of the plane, passing through the three given points:
x x1
y y1
z z1
x2 x1
y2 y1
z2 z1 0
x3 x1
y3 y1
z3 z1
.
(24)
Example. Write an equation of the plane passing through the
M 1 (1;2;3) ; M 2 (3;5;7) ; M 3 (2;4;3) .
By formula (24), we have
x 1
y2
z3
x 1 y 2 z 3
3 1 5 2 7 3 2
2 1 4 2 3 3
3
( x 1)
3 4
( y 2)
2
4
3
6
4
6
3
2
( z 3)
2
3
3 2
26 x 26 ( y 2) 0 13z 39 0 or 2 x z 1 0 .
2
6
The Angle between Planes
Consider the planes given by the equations
A1 x B1 y C1 z D1 0 ; A2 x B2 y C 2 z D2 0 ,
which have normal vectors N 1 A1 ; B1 ; C1 and N 2 A2 ; B 2 ; C 2 . Using inner
product, we find the cosine of the angle:
cos
N1 N 2
N1 N 2
A1 A2 B1 B2 C1C 2
A1 B1 C1 A2 B2 C 2
2
2
2
2
2
2
.
The condition for two planes to be parallel coincides with the condition for
the normal vectors N 1 and
N 2 to be collinear:
A1 B1 C1
.
A2 B2 C 2
The perpendicularity conditioin coincides with the perpendicularity
condition for the vectors N 1 and
N2 :
( N 1 · N 2 )=0,
i.e., A1A2+B1B2+C1C2=0.
Example 1. Show that the following planes are parallel or perpendicular:
2 x 3 y z 15 0; x 2 y 4 z 7 0 .
The planes are perpendicular, because
А1 А2 В1 В2 С1С 2 2 1 3 2 1 4 2 6 4 0 .
Example 2. Write an equation of the plane passing through the point М0(–
1;2; 4) and parallel to the plane 6x-7y+5z+11=0.
The normal vector N 6;7;5 is normal to the required plane also. We have
A(x–x0)+B(y–y0)+C(z–z0)=0,
6(x+1)–7(y–2)+5(z–4)=0; 6x–7y+5z=0.
The normal equation of a plane. Consider a plane. Let us draw the perpendicular
ОР from the origin to this plane. Let ,, and be the angels between this
perpendicular and the coordinate axes x,y, and z, and let OP p . It is required to
write an equation of this plane.
z
P
n 0 M(x,y,z)
0
y
x
Take an arbitrary point M(x;y;z) in the plane and consider the radius-vector
OM x; y; z . The unit vector n 0 on the perpendicular ОР has the coordinates
n 0 ={cos α; cos ; cos }.
For any point М in the plane, the projection of the vector ОМ on the unit vector
n 0 equals р:
рr 0 OM p .
n
Consider the scalar product
ОМ n n
0
0
рr 0 OM p ,
n
or, in coordinate form, x cos y cos z cos p .
Thus, we have obtained the normal equation of the plane:
x cos y cos z cos p 0 .
(25)
The normalizing factor. Consider the plane given by the general equation
Ax+By+Cz+D=0.
It is required to reduce this equation to the normal form (25).
Definition. Number is called the normalizing factor if the equation
multiplied by it is normal.
To find the normalizing factor, we multiply the general equation of the plane by a
number term by term:
Ax+By+Cz+D=0.
This equation is normal if the two normality conditions hold, i.e.,
1. (A)2+(B)2+(C)2=1,
2. D<0.
From the first condition, taking out 2 and extracting the square root, we
find the normalizing factor.
1
A2 B 2 C 2
.
The sign opposite to that of the free term D must be taken.
The Distance from a Point to a Plane
Let us calculate the distance from a point
M 1 ( x1 ; y1 ; z1 ) to a plane
Ax By Cz D 0 .
z
М1
N
K
лллллллл
л
y
0
x
Since point К belongs to the required plane, its coordinates must satisfy the
equation of the plane, and the expression in the parentheses vanishes. Thus, the
distance from the given point to the plane can be calculated by the formula
d
Ax1 By1 Cz1 D
N
Ax1 By1 Cz1 D
A2 B 2 C 2
;
(26)
if the plane is given by a normal equation, then, considering the formula for the
normalizing factor, we obtain
d x1 cos y1 cos z1 cos p .
( 26 )
Example. Find the distance from the point М1(5;2;4) to the plane
x 2 y 2 z 11 0 ,
d
5 4 8 11 12
4.
3
1 4 4
STRAIGHT LINES IN SPACE
The Vector, Parametric, Canonical, and General Equations of a
Straight Line
The position of a straight line in space is determined by a point М 0 (r0 )
on this line and a vector a parallel to the line. Let us write an
equation of such a line in space.
z
M 0 (r 0 ) М (r )
a
0
y
x
To this end, we take an arbitrary point М (r ) on the line, join М0 and М
to the origin, and find the coordinates of the radius-vectors
ОМ 0 r0 x0 ; y0 ; z 0 ,
ОМ r x; y; z.
It is seen from the figure, that M 0 M r r0 .
If the point М belongs to the straight line, then the vectors M 0 M and
collinear.
Consequently, these vectors meet the collinearity condition
a
are
M 0M t a ,
where t is a parameter.
Let us write the collinearity condition in the form
r r0 t a ;
(*)
equation (*) is the vector equation of the given line.
Suppose given the coordinates of the point M0(x0,y0,z0) and the direction vector
a m; n; p. Let us write the left-hand side of equation (*) in the vector form
r r0 ( x x0 )i ( y y 0 ) j ( z z 0 )k
the direction vector is
a mi n j pk .
Let us represent equation (*) in the form
( x x0 )i ( y y0 ) j ( z z 0 )k t (mi n j pk ) .
Equating the respective coefficients of the unit vectors on the right- and left- sides,
we obtain parametric equations of the straight line:
x х0 mt ,
y у0 nt ,
z z pt ,
0
x x0 mt ,
y y0 nt ,
z z pt.
0
or
(27)
Eliminating the parameter t, we obtain the canonical equations of a straight line:
x x0 y y 0 z z 0
.
m
n
p
(28)
Example. Write the canonical equations of the straight line passing through
the point М 0 1;2;0 parallel to the vector a 2;3;5. We compose the
canonical equation by formula (28):
x 1 y 2 z 0
.
2
3
5
Equating each fraction to a parameter t, we obtain the parametric equations of the
line:
x 2t 1,
y 2 3t ,
z 5t.
The general equation of a straight line in space. Since a straight line in space is
represented as the intersection of two planes, the general equation of a straight line
in space has the form of a system
A1 x B1 y C1 z D1 0,
A2 x B2 y C2 z D2 0,
where the first and the second equations are the equations of the corresponding
planes.
It is always possible to transform the general equation of a straight line into a
canonical equation and vice versa.
Since the direction of
follows that
a is perpendicular to those of the vectors N 1 and N 2 , it
i
j
a N1 N 2 A1
B1
A2
B2
i.e., the canonical equation is
k
B
C1 1
B
C2 2
C1
C2
;
A1
A2
C1 A1
;
C 2 A2
B1
,
B2
x x0
B1 C1
B2
C2
y y0
z z0
A1 C1
A1 B1 .
A2
C2
A2
B2
The angle between straight lines in space. The parallelism and perpendicularity
conditions for straight lines. Let us find the angle between intersecting right lines
given by their canonical equations
x x1 y y1 z z1
;
m1
n1
p1
x x2 y y 2 z z 2
.
m2
n2
p2
The angle between these two lines is equal to the angle between their direction
vectors
a2 m2 i n2 j p2 k ,
a1 m1 i n1 j p1 k ;
i.e.,
cos
a1 a2
a1 a 2
m1m2 n1n2 p1 p 2
m1 n1 p1
2
2
2
m2 n 2 p 2
2
2
2
.
The parallelism and perpendicularity conditions for right lines coincide with
the collinearity and perpendicularity conditions of their direction vectors a1 and
a2 .
If straight lines are perpendicular, then
perpendicularity condition is
a1 a 2 , i.e., a1 a 2 0 , and the
m1 m 2 n1 n 2 p1 p 2 0 .
If straight lines are parallel, then the vector a1 is collinear to a 2 , i.e., their
coordinates are proportional, and the proportionality condition is
m1 n1
p
1.
m2 n 2 p 2
The equation of the straight line passing through two given points. Suppose
given two points
z
M 1 ( x1 ; y1 ; z1 ) and M 2 ( x2 ; y2 ; z 2 ) in space.
M2
M1
0
x
y
Let us write the equation of the straight line passing through these points. For the
direction vector we take
a M 1 M 2 x 2 x1 ; y 2 y1 ; z 2 z1 .
The canonical equation (28) of the right line passing through a point М1
parallel to a vector
М 1 М 2 implies
x x1
y y1
z z1
.
x 2 x1 y 2 y1 z 2 z1
(*)
This is the equation of the straight line passing through the points М1 and
М2.
Example. Write the equation of the straight line passing through the points
M 1 (3;5;6) and M 2 (1;2;1) .
Let substitute the given coordinates into (*):
x3 y5 z 6
, whence
1 3 2 5 1 6
x3 y 5 z 6
.
2
7
5
The Intersection of a Plane and a straight Line
Suppose given a plane Ax+By+Сz+D=0
and a straight line
x x0 y y 0 z z 0
m
n
p
in space.
If the straight line and the plane do not intersect, then the parameter t1 does
not exist. Suppose, that the line is parallel to the plane; then the vectors a and N
are perpendicular, i.e., or ( a · N )=0, and Am+Bn+Cp=0. This means that the
denominator vanishes.
N { A; B; C}
a {m; n; p}
Thus, the parallelism condition for a straight line and a plane is
Am+Bn+Cp=0.
Suppose that the line is perpendicular to the plane; then the vectors a and N are
collinear, i.e., their coordinates are proportional:
A B C
.
m n p
This is the perpendicularity condition for a straight line and a plane.
Example. Write the equation of a straight line perpendicular to the plane 2x3y+4z+11=0 and passing through the intersection point of this plane with the line
x 1 y 2 z
.
3
2
1
Let us reduce the canonical equation to the parametric form:
x 3t 1,
y 2t 2,
z t.
Substituting these expressions for the variables x,y, and z into the equation of the
plane
6t+2–6t+6+4t+11=0.
Thus, 4t=–19, t
19
, and the intersection point of the line with the plane has
4
the coordinates
х
57
53
1 ;
4
4
у
38
46
2 ;
4
4
z
19
.
4
The normal vector to the plane is N 2;3;4 ; since the straight line is
perpendicular to the plane, it follows that its direction vector is collinear to the
normal vector, i.e.,
equations
a N . Thus, the required straight line is determined by the
53
46
19
y
z
4
4
4 .
2
3
4
x
The angle between a plane and a straight line. Suppose that a plane in space is
determined by its general equation
Ax+By+Сz+D=0,
and a straight line is determined by its canonical equations
x x0
y y0 z z0
.
m
n
p
2
Take the acute angle between the right
line and its projection on the plane for
the angle between the straight line
and the plane.
а
N
Using the scalar product of vectors
( a N ) and the general scheme, we determine
the angle from the angle between the direction vector a {m; n; p} and the
normal vector
N { A; B; C} ; (a N ) .
2
By the reduction formula, sin cos( a N ), whence
sin
Am Bn Cp
A2 B 2 C 2 m 2 n 2 p 2
.
Definition. A directed interval (or an ordered pair of points) is called a
vector.
В
А
AB
Definition. A vector with coinciding endpoints is called the null vector.
Definition. The distance between the head and tail of a vector is called the
length a , or absolute value of this vector. It is denoted by a or AB .
Definition. Vectors are collinear if they lie on the same straight line or on
parallel lines.
A
B C
a
b
АВ, АС are collinear vectors,
А1
В1
А1 В1 , В1 А1 are collinear vectors.
Definition. Vectors are coplanar if they lie in the same plane or in parallel
planes.
Definition. Two vectors are said to be equal if they are collinear and have
the same direction and length.
Linear Operations on Vectors
Definition. The product of a vector a and a real number is the vector b
defined by the following conditions:
(1) b
a;
(2) The vector b is collinear to a ;
3. The vectors b and a have the same direction if λ>0 and opposite
directions if λ<0. If λ=0, then the direction of the vectors is arbitrary.
а
1
3а
а
2
Property 1. For any numbers α and β and any vector
α(β а )= (αβ) а
а,
Definition. Suppose that AB and CD are vectors and E is a point such
that BE = CD . Then the vector AE is called the sum of the vectors AB and CD
and denoted by AB + CD
Eε
а b
В
а
A
D
а b
b
C
Property 2. Addition of vectors is commutative; this means that, for any
two vectors,
а b =b а .
Property 3. Addition of vectors is associative; this means that, for any
vectors,
a (b c) (a b) c .
Property 4. Addition is distributive with respect to multiplication by a
number; i.e., for any vectors
а and b and any number ,
( a b) a b .
Property 5. For any numbers and and any vector a ,
( )a a a .
Definition. Free vectors are vectors which can be translated, which means
that they do not depend on the head but are determined by direction and length.
B
B1
B
A A1
A
Consider vectors a1 , a 2 ,..., a k , their sum a a1 a 2 ... a k is determined by one
vector, whose head coincides with that of the first vector and the tail with that of
the last vector.
b a1 a 2 ,
c a1 a 2 a 3 .
аk
а3
а
c
b
а2
а1
Definition. The ort-vector of a vector
direction coincides with that of а .
ea
1
e
a
a
а is the vector of unit length whose
is the ort-vector of
а.
а
ea
Subtraction of vectors can be considered as the addition of two vectors, the second
of which is taken with the sign –:
a b a (1)b .
Definition. The projection of a vector onto an axis is defined as the length
of the interval whose endpoints are the projections of the endpoints of the vector
onto this axis which is taken with the sign + if the angle between the vector and
the axis is acute and with this sign – if this angle is obtuse:
prl АВ A1 B1 AB cos .
Decomposition of vectors.
Theorem 1. An arbitrary vector с in the plane can be decomposed into two
noncollinear vectors:
c a b .
Theorem 2. An arbitrary vector d in space can be decomposed into three
noncoplanar vectors
d a b c .
Let a, b, and c be noncoplanar vectors.
The Cartesian system of coordinates. Consider the following coordinate
system: take mutually perpendicular unit vectors i, j and k , draw coordinate axes
x,y and z along them, and fix a unit on metric scale:
z
i j k 1,
M(x1,y1,z1)
k
0
A
i
B
j
i jk.
y
M1(x1,y1,0)
x
Definition. The triple of vectors i, j , k is called right if, looking from the
endpoint of the last vector, we see that the shorter rotation from the first vector to
the second is anticlockwise.
From the triangle ОММ1, we obtain
OM OM 1 M 1 M ,
Since the vector М 1 М is collinear to the unit vector k , it follows that
M 1 M z1 k .
From the triangle ОАМ1, we obtain
OM 1 OA OB x1 i y1 j ,
because, by analogy, the vectors OA and OB are collinear to the unit vectors
and j . Substituting the vector
i
OM 1 thus obtained, we see that
OM x1 i y1 j z1 k .
(2)
Thus, the radius vector OM is represented as the sum of i, j, and k and multiplied
by the corresponding coordinates of the point М.
Consider the vectors
а x1 i y1 j z1 k
and
b x2 i y 2 j z 2 k
and their sum
a b x1 i y1 j z1 k x2 i y2 j z 2 k ( x1 x2 )i ( y1 y2 ) j ( z1 z 2 )k .
Under addition the respective coordinates are added
Let us multiply the vector а by a number :
a x1 i y j z1 k .
When a vector is multiplied by a number , each coordinate of this vector is
multiplied by this number.
Example. Find the vector 4 AB 3CD if
AB 2;3;0;
CD 1;2;4.
Let us find the required vector in vector notation:
4 AB 3CD 4(2i 3 j 0k ) 3(i 2 j 4k ) 11i 18 j 12k ,
4 AB 3CD 11i 18 j 12k .
To find the same vector in vector notation, we multiply the first vector by 4 and the
second by –3 and sum their coordinates:
4 AB 8;12;0
3 CD 3;6;12
4 AB 3 CD 11;18;12.
Given two points M 1 ( x1 ; y1 ; z1 ) and M 2 ( x 2 ; y 2 ; z 2 ) in space, find the
vector M 1 M 2 .
Thus, we have found the required vector in the coordinate notation:
M1M 2 x2 x1; y2 y1; z2 z1 .
(2)
To find the coordinates of a vector, we must subtract the coordinates of its
tail from the coordinates of the head.
For example, let us find vectors with given coordinates of heads and tails:
М1(7;4;–3); М2(1;–2;–2);
М 1 М 2 ={–6; –6; 1};
М 2 М 1 ={6; 6; –1}.
Find the length of a vector
а x1 ; y1 ; z1 :
a ?
a OM 1.
From the right triangle ОМ1М2 , we find the hypotenuse
2
2
2
ОМ 1 OM 2 M 1M 2 ,
z
where
M1M 2 z1 .
From the other right triangle ОАМ2 ,
we find the hypotenuse ОМ 2
Substituting it into
the first hypotenuse, we obtain
ОМ 1
2
2
0
M1
z1
y1
M2
y
x1 2 y1 2 .
А
x
x1
x1 2 y1 2 z1 2 .
Thus, the length of a vector is defined by the formula
a ОМ1 x12 y12 z12 .
(3)
Inner Product of Vectors and its Properties
Definition. The inner product of two vectors а and b is the product of the
absolute values of these vectors and the cosine of the angle between them:
(ab) a b cos(a b) .
(4)
A
0
a
φ
В
b
Property 1. The inner product of two vectors is equal to the product of the
absolute value of one vector and the projection of the second vector onto the first,
i.e.,
(ab) a b cos b рrb а a рra b .
(5)
Property 2. The inner product of two vectors is equal to zero if and only if
these vectors are perpendicular.
Property 3. The inner product of vectors is commutative:
( ab) (b a ).
Property 4. To multiply an inner product by a number , it is sufficient to
multiply one of the factors by :
(a b) ( a b) (a b).
Property 5. Inner product is associative:
(a b) c (a c) (b c) .
The inner product of vectors in coordinates.
Consider two vectors
and b x 2 i y 2 j z 2 k .
Thus, six of the nine terms are zero, and the remaining three terms are
a x1 i y1 j z1 k
(ab) x1 x 2 y1 y 2 z1 z 2 .
The inner product of vectors is equal to the sum of products of their
coordinates.
Example 1. ( ab) -? a 2i 3 j 5k and b 4i 2 j k , then
(ab) (2i 3 j 5k ) (4i 2 j k ) 2 4 2 3 5 1 3 .
Example 2. a {3;1;5} and b {2;4;7} . Then
(ab) 3 2 1 4 5 7 33 .
Let us derive a formula for the length of a vector by using inner product:
2
(a a) a a cos 0 a .
By (6), it equals
a a x1 x1 y1 y1 z1 z1 .
Thus, we obtain
a x12 y12 z12
.
(6)
The direction of a vector. Let us find the angle between two vectors а and b .
Consider the inner product
(ab) a b cos .
We have
cos
( ab)
ab
.
(*)
Writing the product and absolute values in coordinates, we obtain
cos
x1 x 2 y1 y 2 z1 z 2
x1 y1 z1
2
2
x2 y 2 z 2
2
2
Example 3. Find an angle between
b i 2 j 2k . By using formula (**), we find
1 2 8
vectors
2
2
.
a i j 4k
(**)
and
2
, 135
2
1 1 16 1 4 4
Let us determine a condition for vectors to be perpendicular. Suppose that vectors
cos
а and b are perpendicular, i.e.,
2
; then (ab) 0 , and
x1 x 2 y1 y 2 z1 z 2 0 .
This is the condition for vectors to be perpendicular.
z
а
k
i
0
y
j
x
Consider the angles between a vector
denote these angles by
(a i) ;
а and the unit vectors i, j and k . We
(a j ) ;
(a k ) .
Take the product of a and any unit vector, say, i = {1;0;0}
(a i ) x1 1 y1 0 z1 0 x1 .
By formula (*), the cosine of the angle from it is
(7)
cos
x1
ai
x1
a
.
Similarly the cosines of the other angles are
cos
x1
, cos
a
y1
a
, cos
z1
.
a
(8)
These cosines are called the directional cosines of the vector а .
The sum of the squared directional cosines equals one:
cos 2 cos 2 cos 2 1 .
To prove this, it sufficies to square the cosines by formula (8) and sum them:
cos2 cos2 cos2
x1
a
2
2
y1
a
2
2
z1
a
2
2
x1 y1 z1
2
2
a
2
2
a
a
2
2
1.
Example 5. For what are the vectors
a i 3 j 2k
and
b i 2 j 2k
perpendicular?
We use the perpendicularity condition (7) and write the inner product of the given
vectors in coordinates:
x1 x 2 y1 y 2 z1 z 2 0 ;
1 3 2 2 2 0 , =10.
Vector Product and Its Properties
Consider two vectors а and b :
Definition. The vector product of two vectors а and b is a
vector с , satisfying the following conditions:
(1) the absolute value of с equals the product of
the absolute values of the two given vectors and the sine of
the angle between them:
c a b sin( a b) ;
c
b
a
(*)
(2) the vector c is perpendicular to both vectors
а and b :
c a, c b ;
а , b and с constitute a right triple of vectors (that is,
looking from the tail of, we see that the shorter rotation from а to b is carried out
anticlockwise). The vector product of а and b is denoted by
(3) the three vectors
[a b ] c .
1.The absolute value of the vector product of two vectors is equal to the area
of the parallelogram spanned by these vectors:
[a b ] S рar .
2. The vector product is anticommutative, i.e.,
[a b ] [b a ] .
3. To multiply a vector product by a number , it is suffices to multiply one
of the vectors by this number (without proof):
[ab] [ ab] [a b] .
4. Vector product is associative:
[( a b)d ] [a d ] [b d ] .
5. The vector product of collinear vectors is equal to zero, and vice versa, if
the vector product of two vectors is zero, then these vectors are collinear.
The vector product in coordinates.
Consider vectors
a x1 i y1 j z1 k and b x 2 i y 2 j z 2 k .
[ab] ( y1 z2 z1 y2 )i ( x1 z2 z1 x2 ) j ( x1 y2 y1 x2 )k
y1
z1
y2
z2
i
x1
z1
x2
z2
j
x1
y1
x2
y2
k.
Note that the right – hand side is the expansion of a third – order determinant along
the row with elements i, j and k .
Thus, the coordinates of the vector product are determined form the third – order
determinant as
ab
i
j
k
x1
y1
z1
x2
y2
z2
,
(9)
and its absolute value is
[a b ]
y1
z1
y2
z2
2
x1
z1
x2
z2
2
x1
y1
x2
y2
2
.
Example 1. Given the vectors a 3;1;2 and b 1;2;1, find the
vector products (а) [ a b ] ; (b) [( 2a b)( 2a b)]
(а) Let us use the expression (9) a vector product:
i
j
ab 3
k
1 2 i(1) 2
1
1 2
1 2
2
1
j (1) 3
3 2
1 1
k (1) 4
3 1
1 2
5i j 7k ,
[ a b ] {5;1;7} .
(b) Let us find the required product by using associativity:
[( 2a b)( 2a b)] 4[a a] 2[ab] 2[b a] [bb]
2[ab] 2[ab] = 4[ a b ] ,
[( 2a b)( 2a b)] = 4[ a b ] .
Since the second product is expressed linearly in terms of the first, it sufficed to
multiply the coordinates of the first vector by 4:
[ a b ] {5;1;7} , whence
4 : [ab] {20;4;28} [( 2a b)( 2a b)] .
Triple Product of Vectors and its Properties
Definition. The triple product of three vectors is the inner product of the
third vector by the vector product of the first two vectors; it is denoted by
[ a b ] c ( abc ) .
Definition. The vector product of the vector product of the first two vectors
and the third vector ones is called the double vector product:
[ [ab] c ] .
Since double vector product is used very rarely, it have been little studied.
Property 1. The triple product of three vectors equals the volume of a
parallelepiped spanned by these three vectors.
Corollary. It is easy to derive an expression for the volume of a
pyramid from the formula:
S
c
B
b
А
а
С
1
1 1
1
1
Vрyr= Sbase Н = . Sрar Н = Vрar= abc ,
3
3 2
6
6
Vрyr=
1
6
abc .
The sign is needed to obtain a positive volume.
Property 2. Triple product is commutative, and
[ab]c [bc]a [c a ]b .
Property 3. A constant multiplier of any vector can be factored out of
scalar triple product:
[a b]c abc .
Triple
Product
in
Coordinates.
Given
three
vectors
a x1 ; y1 ; z1 , b x 2 ; y 2 ; z 2 , and c x3 ; y3 ; z 3 , let us express the triple
product of these vectors in terms of their coordinates. Consider the triple product
a b c ( [ab] c) .
The vector product equals
a b i yy
1
z1
2
z2
j
x1
z1
x2
z2
k
x1
y1
x2
y2 .
Taking its inner product with c , we obtain
( ab c) x3
y1
z1
y2
z2
y3
x1
z1
x2
z2
z3
x1
y1
x2
y2
;
this is a third – order determinant expanded along the last line, i.e.,
x1
y1
z1
a b c x2
y2
x3
y3
z2 .
z3
Thus, the triple product of three vectors equals the third – order
determinant of the composed of the coordinates of these vectors.
Example 1. Determine the volume of a pyramid ABCD from the coordinates
of its vertices.
D(1;5;2)
B(–1;1;3)
A(1;2;0)
C(0;2;–3)
Compose the vectors
АВ 2;1;3 ,
АС 1;0;3 ,
AD 0;3;2 .
Let us find the volume of a pyramid by the formulas proved above:
Vрyr
2 1 3
1
1
5
abc 1 0 3 4 cube units.
6
6
6
0
3
2
The triple product of coplanar vectors equals zero.
The triple product equals
abc ( [ a b ] c ) (d c) 0 , because d c .
Thus, the coplanarity condition is
x1
y1
z1
x2
y2
z 2 0.
x3
y3
z3
Example 2. Show that the four points А(1;2;–1), В(0;1;5),
D(2;1;3) belong to the same plane.
B
С(–1;2;1), and
Compose the vectors
C
AB ={–1;–1;6},
A
D
AC ={–2;0;2},
AD ={1;–1;4}.
To show that they are coplanar, we find the triple product
1 1 6
( AB AC AD ) 2
1
0
1 1
2 2
1 4
Thus, the four points belong to the same plane.
2
0
0
6
2 0.
2
LECTURE 6.
THE CURVES OF THE SECOND ORDER. CANONICAL EQUATION OF
SECOND ORDER CURVES
LECTURE PLAN:
1. Ellipse
2. Hyperbola
3. Parabola
Ellipse
Definition. An ellipse is the locus of points for which the sum of distances to two
fixed points is constant and equal to 2а.
Take two fixed points at a distance 2c apart, join them by a straight line, and
extend this line to the x-axis. We draw the perpendicular line through the center of
the segment between the focuses and take it for one coordinate axis.
Let us derive the equation of the ellipse.
у
M4(0;b)
М(х;у)
r1
r2
M1(–a;0) F1 (–c;0)
0
F2 (c;0)
M2(a;0)
х
M3(0;–b)
The points F1 and F2 are called the foci of the ellipse, and r1 and r2 are its focal
radii.
To derive the equation of an ellipse, we take an arbitrary point М(х,у) and
consider the distances to the foci:
F1 M r1 ( x c) 2 y 2 ,
F2 M r2 ( x c) 2 y 2 .
The characteristic feature of this line is, by definition,
r1 r2 ( x c) 2 y 2 ( x c) 2 y 2 2a .
This is the equation of the ellipse. Let us reduce it to a convenient form:
( x c ) 2 y 2 4a 2 4a ( x c ) 2 y 2 ( x c ) 2 y 2 ,
x 2 2cx c 2 4a 2 4a ( x c) 2 y 2 x 2 2cx c 2 .
Eliminating some terms and reducing by 4, we obtain
cx a 2 a ( x c) 2 y 2 .
Let us square both sides:
c 2 x 2 2a 2 cx a 4 a 2 [ x 2 2cx c 2 y 2 ],
c 2 x 2 a 4 a 2 x 2 a 2c 2 a 2 y 2 .
We obtain
(c 2 a 2 ) x 2 a 2 y 2 a 2 (c 2 a 2 ) .
2
2
2
Let us divide both sides by (c a )a :
x2
y2
2 2
1 ;
a c a2
changing the sign, we obtain the equation of the ellipse:
x2
y2
1.
a2 a2 c2
Since the length 2a of a polygonal line is larger than the length 2c of a straight line,
we can denote the difference of squares by
a2 c2 b2 .
(*)
Thus, we obtain the classical equation of an ellipse:
x2 y2
1.
a 2 b2
(17)
Intersection points of an ellipse with the coordinates axes. To find the
intersection points of an ellipse with the x-axis, we must solve the system of
equations
y 0,
y 0,
Ox : x 2 y 2
1, x 2 a 2 x a.
a 2 b2
We obtain two vertices of the ellipse: М1(–а;0),
М2(а;0),
M 1 M 2 2a is called the major axis of the ellipse;
а is the major semi axis
We find the intersection of the ellipse with the y-axis by solving the system
x 0,
x 0,
Oy : x 2 y 2
2
2
a 2 b2 1, y b y b.
We obtain the two other vertices of the ellipse, М3(0;–b) and М4(0;b).
M 3 M 4 2b is called the minor axis of the ellipse, and
b is the minor semi axis.
It is seen from equation (17) and the figure that the ellipse is symmetric with
respect to the axes Ox and Oy .
The eccentricity and directrix of an ellipse. Consider the focal radii of an ellipse
r1 ( x c) 2 y 2 ;
r2 ( x c) 2 y 2 .
By definition, we have
r1 r2 2a .
Consider the difference of squares
2
2
r2 r1 x 2 2cx c 2 y 2 x 2 2cx c 2 y 2 ;
2
2
r1 r2 4cx ,
(r1 r2 )(r1 r2 ) 4cx .
or
y
М(х;у) d2
d1
r1
F1(–c;0) 0
x=–l
r2
F2(c;0)
x
x=l
to determine the focal radii, we solve the system of equations
2c
r
r
x,
1 2
a
r1 r2 2a,
c
2
r
2
a
2
x,
1
a
or
c
2r2 2a 2 x,
a
c
r
a
x,
1
a
r a c x.
2
a
Definition. The ratio of distances between the foci to the sum of focal radii
is called eccentricity:
2с с
.
2а а
If the distance between the foci is less than 2а, then the eccentricity is
Thus, the focal radii of the ellipse are
с
1.
а
r1 a x ,
r2 a x .
Definition. The directrix of an ellipse is the straight line parallel to the y-axis
such that the ratio of the focal radius to the distance from an ellipse point to it is
constant and equal the eccentricity.
Let us draw two straight lines x=–l and x=l parallel to the y-axis and find l
such that the ratio of the focal radius to the distance from a point М to this straight
line is constant and equals the eccentricity:
r2
.
d2
Substituting the distance and the focal radius, we obtain
r2 a x
d2
lx
a
x
The ratio is equal to the eccentricity when l
lx
a
.
, i.e., х
analogy, we obtain equations of the directrices:
a
х ;
х
a
,
r2
r1
1.
where
d 2 d1
Tangents to an Ellipse
Consider the equation of an ellipse:
x2 y2
2 1.
2
a
b
у
М0(х0;у0)
0
х
a
is the directrix. By
As is known, the equation of a tangent to a curve is determined by the formula
y y0 y ( x0 )( x x0 ) .
Differentiating the equation of the ellipse as an implicit function, we obtain
b 2 x0
2 x 2 y y
b2 x
0 , whence y 2 , or k 2 .
2
2
a y0
a y
a
b
Substituting this k, we find the equation of the tangent line:
b 2 x0
y y0 2 ( x x0 ) .
a y0
Let us transform it:
2
2
a 2 yy0 a 2 y0 b 2 x0 x b 2 x0 ,
2
2
a 2 yy0 b 2 xx0 b 2 x0 a 2 y0 . Dividing by a 2b 2 , we obtain
2
2
xx0 yy0 x0
y0
.
a2
b2
a2 b2
Since the point М0 belongs to the ellipse, the coordinates of М0 must satisfy its
equation, and the right-hand side equals one.
Thus, the equation of a tangent to an ellipse is
xx0 yy0
2 1.
(18)
a2
b
2
2
Example. Given the ellipse given 16 x 25 y 400 , find the distance
between its foci, eccentricity, and the equations of directrices.
Let us reduce the equation it to the classical form (17):
x2 y2
1 ; a 2 25; b 2 16 ;
25 16
c 2 25 16 9, c 3 , 2c 6 .
Let us find the eccentricity:
c 3
.
a 5
The equations of the directrices are
x
a
,
x
25
1
8 .
3
3
Hyperbola
Definition. The locus of the points for which the difference of distances to two
fixed points is constant equal to 2а is called a hyperbola.
As for an ellipse, we introduce a new coordinate system:
у
М(х,у)
r1
r2
F1(–c,0) M1
M2
F2(c,0) x
To derive the equation of a hyperbola, we take an arbitrary point М(х,у) on the
hyperbola and consider the distances from this point to the foci:
F1M r1 ( x c) 2 y 2 ; F2 M r2 ( x c) 2 y 2 .
The characteristic feature of the line is, by definition,
( x c ) 2 y 2 ( x c ) 2 y 2 2a .
We have composed an equation of the hyperbola. Let us reduce to a convenient
form (by analogy with the ellipse):
(c 2 a 2 ) x 2 a 2 y 2 a 2 (c 2 a 2 ) .
We divide both sides by (c a )a :
2
2
2
x2
y2
2 2
1
a c a2
Changing the sign, we obtain the equation of a hyperbola:
x2
y2
1.
a2 c2 a2
Since 2a<2c, we denote the difference of squares by
c2 a2 b2 .
(**)
Thus, we have obtained the classical equation of a hyperbola:
x2 y2
2 1.
2
a
b
(19)
Intersection points of a hyperbola with the coordinates axes. To find the
intersection points of a hyperbola with the x-axis, we must solve the system of
equations
y 0,
y 0,
Ox : x 2 y 2
2
2
a 2 b2 1, x a x a.
We obtain the vertices of the hyperbola М1(–а;0) and М2(а;0). The distance
M 1 M 2 2a is called the real axis of a hyperbola and а is called its real
semiaxis.
We find the intersection of a hyperbola with the y-axis by solving the system
x 0,
x 0,
Oy : x 2 y 2
1, y 2 b2 .
a 2 b2
This system has no solution, i.e., the hyperbola does not intersect the
y-axis; for
this reason,
2b is called the imaginary axis of a hyperbola, and
b is the imaginary semiaxis.
It is seen from the equation and the figure that the hyperbola is symmetric with
respect to the axes Ox and Oy .
We determine у from the equation of a hyperbola:
y2 x2
1 ;
b2 a 2
As
y
b 2
x a2 .
a
x , y . Consequently, the hyperbola is unbounded.
Asymptotes of a Hyperbola
Definition. The asymptote of a curve is a straight line approached by the curve
line at infinity.
An oblique asymptote is determined by an equation of the form an oblique
y=kx+b.
To find the slope, we suppose that b=0; then
k lim
x
We find b from the equation
y
k lim
x
x ;
b y kx .
b 2
x a2
b
a
.
x
a
Passing to the limit, we obtain
b x 2 a 2 x 2 ab
b
b 2 2 b
b lim ( y kx) lim y x lim x a x lim
0.
2
2
x
x
x a
x a
a
a
x a x
Thus, the hyperbola has two asymptotes, passing through the origin:
y
b
x;
a
y
b
x.
a
First, we construct a rectangle with sides 2а along the х-axis and 2b along the уaxis. We draw the diagonals in this rectangle and extend them; they are the
asymptotes of the hyperbola.
y
y
b
x
a
2a 2b
F1 М1(–а;0)
М2(а;0) F2
0
x
From the vertices M1(–a;0) and M2(a;0), we draw the branches of the hyperbola
approaching the asymptotes.
The directrices of a hyperbola. By definition, r2 r1 2a . By analogy with an
ellipse, we find the difference of squares
2
2
r2 r1 4сх ,
(r2 r1 )(r2 r1 ) 4cx .
As a result, we obtain
r2 r1 2a,
cx
r2 r1 2 a ,
r2 a
c
x
a ;
r1 a
c
x.
a
By definition,
c
a
c
1 ,
is the eccentricity of the hyperbola; we have
a
because с>а.
By analogy with an ellipse, the directrices of a hyperbola are determined by the
a
a
x
equations
and x .
M1(–а;0)
x
x
0
M2(а;0)
a
x
a
Since the eccentricity of a hyperbola is
x
a
y
1 , it follows that the directrix is
c ; i.e., the directrices lie between the two vertices.
The ratios of the focal radii to the corresponding distances from an arbitrary point
of the hyperbola to the directrices is constant and equals the eccentricity:
r2
r
1 1.
d 2 d1
Tangent lines to a hyperbola. Suppose given, the equation of a hyperbola and a
point М0(х0,у0) on it:
x2 y2
1.
a 2 b2
As is known, the equation of a tangent to a curve is
y y0 y ( x0 )( x x0 ) .
Let us differentiate the equation of a hyperbola as an implicit function:
2 x 2 y y
0;
a2
b2
we obtain
b2 x
y 2 , or
a y
Substituting, we obtain the equation of the tangent line:
b 2 x0
y y 0 2 ( x x0 ) ,
a y0
b 2 x0
k 2 .
a y0
Let us transform it:
2
2
a 2 yy0 a 2 y0 b 2 x0 x b 2 x0 ;
2
2
a 2 yy0 b 2 xx0 b 2 x0 a 2 y0 . Dividing by a 2b 2 , we obtain
2
2
xx
yy
x
y
20 20 02 02 .
a
b
a
b
Since the point М0 belongs to the hyperbola, its coordinates must satisfy the
equation of the hyperbola, and, the right-hand side equals (–1) .
Thus, the equation of a tangent to a hyperbola is
xx0 yy 0
2 1.
a2
b
Example. Write the equation of the hyperbola with real semi axis 4 and foci
at the points F1(–5;0) and F2(5;0).
Formula (19) gives
x2 y2
1
a 2 b2
We have a 2 16 and c 2 25 ; thus b 2 c 2 – a 2 =25–16=9.
The required classical equation of the hyperbola is
x2 y2
1.
16 9
Parabola
Definition. The locus of points for which the distance to a fixed point equals
the distance to a given straight line (a directrix) is called a parabola.
Let us draw the perpendicular through a fixed point to the given straight line
and take it for the x-axis. From the middle point of the segment joining the focus to
the given straight line we draw a perpendicular and take it for the y-axis.
y
p
N ; y
2
p
М(х;у)
r
0
p
F ;0
2
x
To derive the equation of the parabola, we take an arbitrary point М(х;у) on it and
write down the characteristic feature of a parabola as a mathematical formula.
The distance from the focus to the directrex is called the parameter of the
parabola and denoted by p. Let us find the distance from the point М(х;у) to the
focus:
2
p
r x y2 ,
2
2
p
MN = d x 0 .
2
and
By definition, these distances are equal:
2
2
p
p
2
x y x .
2
2
Let us transform this, relation by squaring both sides:
2
2
p
p
2
x y x ;
2
2
p2
p2
2
2
x xp
y x xp
.
4
4
2
We obtain
y 2 2 px .
(20)
This is the classical equation of a parabola.
The parabola passes through the origin (0;0), because it satisfies equation (20).
Suppose that the parameter is a positive number р>0; then, since у2>0, we
have x>0, and the parabola is contained in the right half-plane. If p<0, then x<0,
and the parabola is contained in the left half-plane
y
у
p>0
p<0
0
x
0
х
M0(x0,y0)
Consider the equation of a parabola in the “school” form x 2 py . Let us
analyze this equation by analogy with (20): if p>0, then y>0, and the branches of
the parabola are directed upward; if p<0, then y<0, and the branches of the
parabola are directed downward.
2
p>0
y
p<0
y
0
0
x
x
The eccentricity of the parabola, that is, the ratio of the focal radius to the distance
from a point to the directrix, equals 1, i.e.,
r
1.
d
Tangent lines to a parabola. Given a point М0(х0,у0) on a parabola, it is
required to write the equation of a tangent to the parabola at this point.
Let us find the slope of the tangent:
k y ( x0 , y0 ) ,
To this end, we differentiate equation (20) as an implicit function:
p
p
y x , or k .
y0
y
2 yy 2 p , whence
Substituting this into the equation of a straight line with given slope, we obtain
y y0
p
( x x0 ) ;
y0
2
yy0 y0 px px0 .
Since the point М0 (х0;у0) belongs to the parabola, its coordinates satisfy the
equation of the parabola:
2
y0 2 px0 ; or
yy0 px 2 px0 px0 .
Thus, we obtain the equation of a tangent to the parabola
yy0 p( x x0 ) .
Example. Write the classical equation of the parabola with directrix х=–5 .
The parabola is given by the equation
y 2 2 px , and the directrix by the
p
p
5 and р=10.
x
equation
2 , which means that 2
Then, the required equation of a parabola is y 2 20x .
Definition. The locus of points for which the ratio of the distances to focal
radii to the distances to the corresponding directrices is constant and equal to the
eccentricity
r1
r
r
2 , which is
d1 d 2 d
(1) less than 1, then it is called an ellipse;
(2) larger than 1, is called a hyperbola;
(3) equal to 1, is called a parabola.
A General Equation of a Second-Order Curve and Its Classical Form
Definition. An equation of the form
Ax 2 2Bxy Cy 2 2Dx 2Ey F 0
(21)
is called a general equation of a second-order curve.
Assigning particular values to the coefficients, we obtain the above equations of a
hyperbola, an ellipse, and a parabola.
LECTURE 5
FUNCTION. FUNCTION LIMIT. FUNDAMENTAL THEOREMS ON
LIMITS. INFINITELY SMALL AND INFINITELY LARGE QUANTITIES.
THE ENDS
LECTURE PLAN:
1. Function limit
2. Infinitely small and infinitely large quantities
3. Main theorems about limits and their applications
4. Continuity of function
The Theory of Limits
The limit of a sequence.
Definition. A sequence is an infinite set of terms, each of which is assigned
a number. The terms of a sequence must obey a certain law.
х1,х2,х3,х4,…,хn,….
1
х
;
n
Example.
3n
1 1 1
1
; 2 ; 3 ;... n ;... .
3 3 3
3
Definition. A number а is called the limit of a sequence if, for any
there exists a number N( ) depending on , such
хn а
Notation:
for n>N,
lim xn a .
n
>0,
Example.
xn
1 1
1
1
1
1
, 0 , , n , N ,
, N 100, n 100 .
n n
n
100
Definition. The limit of a variable х is a number а such that for any
there exists an х starting with which all х satisfy the inequalities
>0,
x a , x a .
Notation:
lim x a, x a.
а–
(а– ;
)
Properties:
1. The limit of a constant number equals this number.
2. A variable can not have two different limits.
3. Some variables have no limit.
Example.
0;
xn
1 (1) n
n3
. Assigning integer values to n, we obtain:
2
2
2
;
0
;
;
0
;
etc.; i.e. this variable has no limit.
23
43
63
Definition. We say that х tends to infinity if, for any number М, there exists
an х such that, starting it,
x M , lim х .
x
(a) M>0, x>M, x
;
(b) M<0, x<–M, x – .
Example. xn=n2+1 ; as n
, хn tends to infinity.
The limit of a function. Suppose that y=f(x) is a function defined on a
domain D containing a point а:
D.
Definition. A number b is called the limit of the function f(x) as
а if, for
any given >0, there exists a small positive
depending on
( ( )>0) such
that, for any х satisfying the inequality
х а , f ( x) b . Notation:
lim f ( x) b .
(1)
x a
Example. Find the limit of f(x)=5x–1 as x 2, and determine
.
lim f ( x) lim (5x 1) 9 ,
x 2
x 2
f ( x) b , x 2 , 5 x 1 9 , 5 x 10 ,10 5 x 10 ,
2 x 2 ,2 x 2 ,
5
5
i.e.,
5
.
To find , we must find x from the inequality for the function and substitute it in
the inequality for the variable.
Definition. The left limit of a function f(x) as x a is the limit of f(x) as x
a, and х<а. Notation:
lim f ( x) b .
x a 0
x
Definition. The right limit of a function f(x) as x
a, and х>а. Notation:
a, is the limit of f(x) as
lim f ( x) b .
xa 0
If the left limit equals the right limit and some number b, then b is the limit of the
function as x a.
Example.
99 0
x2 9
f ( x)
, x 3 ; f (3)
an indeterminacy, although the limit
x3
33 0
exists:
( x 3) ( x 3)
x2 9
lim
lim
6.
x 3 x 3
x 3
x3
Definition. A number b is called the limit of f(x) as
there exists a (large) number N depending on
such that
if, for any
>0,
f ( x) b for any
x N.
Notation:
lim f ( x) b .
х
Infinitesimals and bounded functions.
Definition. A function f(x) is said to be infinitesimals as
а if, for any М,
there exists a
such that f ( x) M whenever x а .
Notation:
lim f ( x) .
xa
Definitions. 1. A function f(x) is said to be bounded on a domain D if, for
any х from D,
|f(x)| M.
For example, f(x)=cosx , |cosx| 1, M=1,
y
2
; f ( x) 2 .
1 x2
If this condition is not satisfied, then the function is said to be unbounded, i.e., this
function is an infinite quantity.
2. A function f(x) is said to be bounded as
а if, for any х from a
neighborhood of а, |f(x)| M.
3. A function f(x) is said to be bounded as х
if, for any х>N, |f(x)| M.
Theorem I. If a function f(x) has a finite limit as x a then f(x) is bounded as
x a.
Theorem II. If a function has a limit as
а and this limit is not equal to
zero, then
1
f ( x)
is bounded as
а.
The infinitesimals and their properties.
Definition 1. A function (х) is called an infinitesimal as
lim ( x) 0 .
а if
xa
Definition 2. A function
( а- ; а+ ),
(х) is called an infinitesimal as
а if, for all
(x) .
These two definitions are equivalent, i.e., we can obtain the second
definition from the first and vice versa. (Prove this).
Theorem I. If a function f(x) is represented as the sum of a constant number
and an infinitesimal, i.e.,
f(x)=b+ (x),
(2)
then it has a limit:
lim f ( x) b .
xa
Conversely, if a function f(x) has limit b, then the function can be represented in
the form (2).
Theorem II. If
1
а, then
( х) is an infinite
(х) is an infinitesimal as
quantity.
Theorem III. The sum of finitely many infinitesimals is an infinitesimal:
1
2
к
3
.
Theorem IV. The product (x) z(x) of an infinitesimal (х) by a bounded
function z(x) as x a is an infinitesimal.
Corollary. The product of infinitesimals is an even smaller quantity.
Theorem V. An infinitesimal divided by a function having nonzero limit as
а is infinitesimal, i.e., if
lim v( x) b 0 , then
xa
( х)
v( x)
is an infinitesimal.
Fundamental Theorems on Limits
Theorem I. The limit of the algebraic sum of finitely many functions equals
the sum of the limits of these functions:
lim u1 ( x) u 2 ( x) ... u k ( x) lim u1 ( x) lim u 2 ( x) ... lim u k ( x)
xa
xa
xa
xa
.
Theorem II. The limit of the product of two functions equals the product of
the limits of these functions:
lim u1 ( x) u 2 ( x) lim u1 ( x) lim u 2 ( x) .
xa
xa
xa
Theorem III. The limit of the ratio of two functions equals the quotient of
the limits of the numerator and the denominator:
lim u ( x)
u ( x) x a
lim
.
x a v( x)
lim v( x)
x a
Computations of limits. Examples.
I. Limits as x
.
2 1
2
5x 2 x 1
x
x3
lim
lim
(1) x x 3 x 2 27 x 1019 x
1 27 1019
1
x x2
x3
3
5
2
1
lim
5
x x 2
x x3
5.
1
27
1019 1
1 lim lim
lim
x x
x x
x x3
2
5 lim
The limits in the numerator and the denominator equal zero.
To find the limit of a linear-fractional function, we must divide the
numerator and the denominator by х to the maximum power among the powers of
x in the numerator and the denominator.
1
2
1
2
x2 x 1
x 3 x 4 0 0,
lim
lim x
(2) x
2
x4 2
x
1
1
x4
because х4 is the maximum power of x in the numerator and the denominator.
7
2
x 7
1
x
lim
lim
(3) x x 1
(divide by х2).
x 1 1
0
x x2
1
2
A simple method for finding limits of linear-fractional functions as х
is
to leave the term containing the maximum power of х in the numerator and the
denominator:
x7 x5 1
x7
1
lim
lim
4) x
2 x 7 x 6 5 x 2 x 7 2 ,
5) lim
x
3
3
x3 2 x 1
x3 1
lim
,
x 8 x
8 x 27
8
x 4 3x 2 1
x4
1
lim
lim
6) x
2 x 4 x 3 15 x 2 x 4 2 .
Let us find limits (1), (2), (3) by the simple method:
5x3 2 x 1
5x3
lim
5,
x x 3 x 2 27 x 1019
x x 3
lim
x2 x 1
x2
1
lim
lim
0,
x
x x 4
x x 2
x4 2
lim
x2 7
x2
lim
lim x .
x x 1
x x
x
lim
Deleting the terms containing lower powers of x from the numerator and the
denominator is only possible because, after division by х to the maximum power,
the limits of all such terms vanish.
II. Limits as х а. Looking for a limit, first, substitute x a in the
function. If we obtain a number, then this number is the limit of the function. If we
0
obtain one of the indeterminacies , ,1 , and , then we must eliminate it
0
by transforming the function and then to pass to the limit.
x 3
23
1
lim
(1) x2 3
x 27 8 27 19 ,
x 2 3x 4 0
( x 4)( x 1)
lim
lim
lim ( x 1) 5 ,
(2) x4
x 4
x4
0 x4 ( x 4)
x 1 0
x 1
1
lim
lim
(3) x 1 3
x 1 0 x 1 ( x 1)( x 2 x 1) 3 ,
The First Remarkable Limit and Its Generalization
The following limit exists and equals 1:
sin x
1.
x0 x
lim
Example:
(1)
sin 2 x 0
2 sin x cos x
sin x
lim
2 lim
lim cos x 2 1 1 2 ,
x0
x 0
x
0 x0
x
x x 0
tan x 0
sin x
1
lim
lim
lim
1 1 1 .
2) x 0
x
0
x
0
x
0
x
cos x
lim
The first generalized remarkable limit. The first remarkable limit can be
generalized, namely, written in the more general form
sin ( x)
1.
( x ) 0 ( x )
lim
(4)
x a
In this formula, (х) is an infinitesimal; it is very important that the
argument of the sine and the denominator must be absolutely identical.
Examples.
sin( x 3 27)
1,
(1) lim
x3
x 3 27
(2) xlim
19
sin lg( 20 x)
1.
lg( 20 x)
The Second Remarkable Limit
Consider the limit
1
lim (1 ) n e .
n
n
(5)
The number е satisfies the inequalities 2<e<3, e is an abbreviation for
exponentials, i.e., «outer»; it is sometimes denoted by e=exp and approximately
equals e 2,71828. y=ex is the exponential function.
Examples. Find the following limits by using the second remarkable limit:
x 1
3x 4 3
lim
x 3 x 2
x 1
3x 2 6 3
1 lim
x 3 x 2
6
lim 1
x
3x 2
3x 2
6 x 1
6 3x 2 3
6 ( x 1)
lim
e x 3( 3 x 2 )
e
6
9
e
2
3
.
The Second Generalized Remarkable Limit
The second remarkable limit (*) and its modification (**)
generalized, i.e., written in the more general forms
can be
1
lim 1
N ( x )
N ( x)
x a
N ( x)
e
and
lim 1 ( x)
1
( x)
( x ) 0
e
x a
.
(6)
In these formulas, (х) is an infinitesimal and N(x) is an infinitude. It is very
important that in these formulas, N(x) and (х) are absolutely identical in the
denominators and exponents.
For example,
1
x 2
1
x
1
1 х
1 1
1
2x
lim ln
lim ln
lim ln1
x0 x
x
0
x
0
1 х
x 1 x
2
1 x
1
2x
ln lim 1
2 x 0 1 x
1 x 2 x 1
2 x 1 x x
2x
lim
1
1
1
x 0 x (1 x )
ln e
ln e 2 2 1.
2
2
2
Other Remarkable Limits
Consider the following limits of functions often encountered in applications:
log a (1 x)
log a e ,
x 0
x
ax 1
ex 1
lim
ln a for a=e, lim
1.
x 0
x 0
x
x
(1 x) 1
lim
.
x 0
x
lim
LECTURE 6
THE DERIVATIVE OF THE FUNCTION. GEOMETRIC AND
MECHANICAL MEANING. TABLE OF DERIVATIVES. THE
DIFFERENTIAL OF A FUNCTION
LECTURE PLAN:
1. The derivative of a function
2. Differential of a function
(7)
(8)
(9)
Derivative of Functions
Suppose that there is y=f(x) on the interval [a;b]. We increment х to the
argument х; then the function obtains the following increment
уу+у,
у+у=f(х+х),
у=f(х+х)–у.
у
f(x+x)
f(x)
0
x
x+x
х
Definition. The ratio limit of a function increment to an argument
increment, where the latter tends to zero, is called a derivative of function.
Notation:
y
f ( x x) f ( x)
lim
.
x0 x
x0
x
f ( x) lim
(1)
The notations in the form of f(x), у or ух are the Lagrange notations, and the
notations of the form
dy
df ( x)
or
are the Leibniz notations.
dx
dx
The geometric and mechanical interpretation of a derivative
The mechanistic interpretation of a derivative, i.e., the derivative of
distance with respect to time, is the velocity at a moment t.
The geometric sense of a derivative is that the derivative at the point М0 equals a
slope of a tangent at this point to ОХ -axis.
The function differentiability
Definition 1. Obtaining an assigned function is said to be a function
differentiation.
Suppose that there is у=f(x) given on [a;b].
Definition 2. Suppose that there is the derivative at the point х=х0:
f ( x 0 ) lim
x 0
f ( x 0 x) f ( x 0 )
,
x
then f(x) is said to be differentiable at the point х0.
If f(x) is differentiable at any point of [a;b], the function is said to be
differentiable on the interval.
The derivative of constant, sum, difference,
product and fraction of functions.
1. The derivative of a constant is
с 0 .
1. The derivative of sum (difference) of two functions equals the sum
(difference) of the derivatives of functions:
(uv)= uv.
3. The constant factor can be factored outside the sign of the derivative
(k·u)= k(u).
4. The derivative of the product of two functions equals the derivative of the
first function, multiplied by the second, plus the derivative of the second,
multiplied by the first, i.e.,
(u· v)= uv+uv.
(2)
Example. Find the derivative:
y x 2 1 ln x .
1
y x 2 1 ln x x 2 1(ln x) 2 x ln x ( x 2 1) .
x
5. The derivative of fraction is defined by the following formula
u u v uv
.
v
v2
(3)
1
x
x
3
ln
3
x
3
3 x (3 x ) x ( x) 3 x
2 x
Example.
.
x
х
x
Differentiation of Composite Function
Suppose that there is a composite function denoted by the two-component
chain y=F(u); u=(x), each of these functions is differentiable F(u), (x). Find
the derivative of the composite function F[ (x)].
Theorem. If functions F(u)and (x) are differentiable with respect to their
variables, then there is a derivative of composite function. It is defined as follows:
y x yu u x .
Examples.1. Find the derivative y=cos5x.
Expand the composite function into the chain of the elementary
y=u5; u=cos x.
(4)
Using formula (4), we obtain
yх=5u4 ·(–sin x).
Coming back to our "old" variable, we obtain yх=-5 cos4x sinx.
Table of Principal Derived Functions
x х
а a
х
1
,
e e ,
ln a,
x
log a x
x
ln x 1 ,
1
,
x ln a
x
cos x sin x,
sin x cos x,
tan x
cot x
1
,
2
cos x
arcsin x
arctgx
x
arccos x
1
,
2
1 x
1
,
2
sin x
1
,
2
1 x
arcctgx
1
,
2
1 x
1
,
2
1 x
Differentiation of the Function Given Parametrically
Suppose that there is function y(х) given parametrically:
x (t ),
y (t ).
Suppose that functions (t) and (t) are differentiable in parameter t and
t0, there is also the inverse function t=t-1(x). Then the derivative of function can
be obtained by the formula:
y х
Examples. 1. Find the derivative:
x a cos t ,
y b sin t.
2. Find the derivative:
y x
yt
.
xt
y t
b cos t
b
cot t .
x t a sin t
a
(6)
1
4 (e t 2 1)
t 2 2t
5
ln
e
y ln5 (et 2 1),
2
y
et 1
y x t
.
2 5t.
x
2
cos
5
t
(
sin
5
t
)
5
x
cos
t
Method of Logarithmic Differentiation
The conception of this method is the following: beforehand, we obtain the
logarithm of the assigned function, and only then differentiate the result.
In applications more often we encounter with two cases with the logarithmic
differentiation.
1. Find the derivative of the product of several functions
y=u1(х.)· u2.(х)· u3.(х)· …·un(х).
Let us obtain the logarithm of both parts, using the property of the logarithm of
product
ln y = ln u1.+ ln u2 + ln u3 +…+ ln un.
Differentiate both parts, the left one - as the implicit function
u
u u
1
y x 1 2 ... n .
y
u1 u 2
un
Multiplying both parts by у and substituting it by the function itself, we obtain
u u
u u
u
u
y x y 1 2 ... n u1(х). u2.(х) ….un(х) 1 2 ... n .
un
un
u1 u 2
u1 u 2
Example 1. Find the variable
e x 2 1 x 2 sin 3 x
y
.
x2 x2 1
Find the logarithm of both parts
1
1
ln y x 2 ln(1 x 2 ) 3 ln sin x 2 ln x ln( x 2 1) .
2
2
Differentiate both parts
1
2x
cos x 2
1 2x
y 2x
3
y
2(1 x 2 )
sin x x 2( x 2 1) .
Then, we obtain the required derivative
1
2x
cos x 2
1 2x
y 2x
3
.
y
2(1 x 2 )
sin x x 2( x 2 1)
Then out of this, we obtain the required derivative
e x 2 1 x 2 sin 3 x
x
2
х
y
2
x
3
ctgx
.
1 x2
x 1 x 2
x2 x2 1
2. Consider the function
y=u(x)v(x),
here the base u(x) and v(x) the power are functions; thus, the function is called a
power-exponential function.
Obtain the derivative of this function. By the analogy with the first case, obtain the
logarithm and use the property of the logarithm of power
ln y = v(x). ln u(x).
Differentiate both parts
1
1
y x v ln u u v ,
y
u
multiplying both parts by у and substituting by the power-exponential function;
thus, we obtain the derivative
v
y x u v v ln u u .
u
(7)
Example 2. Find the derivative
y=xx , ln y = x.ln x,
y x
1
ln x x .
y
x
Thus, the required derivtive is
yx x x 1 ln x .
Differential of Function
Suppose that there is the differentiable function y=f(x) on the interval [a;b].
It follows that
y
.
x 0 x
f ( х) lim
Acodrding to the principal theorem on limits, we have
y
f ( x) (х) , (х) 0 as х0 .
x
Find the increment
y f ( x) х (х) х .
(*)
Definition. The principal part of the increment of function (*) is called a
differential of function and is denoted by
dy f ( x) x .
Find the differential of function у=х by definition
dy dx ( x) x
or
dx x ,
i.e., the differential of the independent variable equals the increment of this
variable. Substituting increment х by dх in the ratio, we obtain
dy f ( x) dx .
(8)
The differential of function equals the derivative of this function, multiplied
by the differential of argument.
For example, y = cos2x, dy = –2 cos x·sin x·dx.
As the differential is the derivative multiplied by the differential of argument, then
the differential of function has absolutely all properties, which the derivatives
have, i.e.,
d(uv) = dudv,
d(u·v) = vdu+udv,
u vdu udv
d
.
v
v2
Higher Derivatives
Suppose that there is the differentiable function y=f(x), i.e., there is the
derivative y=f(x), which is denoted by f1(x) .
Definition. The derivative of the derivative of function is called the second
derivative; the derivative of the second derivative of function is called the
derivative of the third order respectively and is denoted by
y x x y x ,
yx x yx ,
2
2
y
3
.......
( n1)
x n 1
x
y x( nn )
etc.
Notations:
dy
y ,
dx
y x'' 2
d 17 y
d2y
(17 )
, etc., y x17
dx17
dx 2
are read, for example, the third derivative – d three y with respect to the third
power of dx.
LECTURE 7
ROLLE'S THEOREM, LAGRANGE, CAUCHY. L'HOPITAL'S RULE
LECTURE PLAN:
1. Properties of differentiable functions
2. Limits- indeterminate forms and L’Hospital’s Rule
PROPERTIES OF DIFFERENTIABLE FUNCTIONS
Fermat's Theorem.
Let y f (x) be defined and differentiable on an open interval (a, b). If f (x) attains
its absolute maximum or absolute minimum (both are called absolute extremum)
at x c , where c (a, b) , then f ' (c) 0 .
Rolle's Theorem.
If a function f (x) satisfies all the following three conditions:
f (x) is continuous on the closed interval [a, b] ,
(1)
f (x) is differentiable in the open interval (a, b) ,
(2)
f (a ) f (b) ;
(3)
then there exists at least a point (a, b) such that f ' ( ) 0 .
Lagrange's Theorem.
If a function f (x) is
(1) continuous on the closed interval [a, b] and
(2) differentiable in the open interval (a, b) ,
then there exists at least a point (a, b) such that
f (b) f (a)
f ' ( ) .
ba
Limits- indeterminate forms and L’Hospital’s Rule
I.
Indeterminate Form of the Type
0
0
We have previously studied limits with the indeterminate form
0
as shown in the
0
following examples:
x2 4
( x 2)( x 2)
Example 5: lim
lim
lim x 2 2 2 4
x 2 x 2
x 2
x2
x 2
However, there is a general, systematic method for determining limits with the
0
. Suppose that f and g are differentiable functions at x = a
0
0
f ( x)
and that lim
is an indeterminate form of the type ; that is, lim f ( x) 0 and
0
x a g ( x )
x a
lim g ( x) 0 . Since f and g are differentiable functions at x = a, then f and g are
indeterminate form
xa
continuous at x = a; that is, f (a) lim f ( x) = 0 and g (a) lim g ( x) = 0. Furthermore,
xa
xa
since f and g are differentiable functions at x = a, then f (a) lim
xa
g ( x) g (a )
g (a ) lim
. Thus, if g (a) 0 , then
xa
xa
f ( x) f (a)
and
xa
f ( x) f (a)
f ( x)
f ( x) f (a )
f (a)
f ( x)
xa
if f and
lim
lim
lim
lim
x a g ( x)
x a g ( x) g (a )
x a g ( x ) g ( a )
g (a) xa g ( x)
xa
g are continuous at x = a. This illustrates a special case of the technique known
as
L’Hospital’s Rule.
The Bernoulli-L'Hopital Rule
This rule is used for finding the ratio limits of the form
0
,
0
Theorem 1. Suppose that there are differentiable functions f(x) and (x) on
f ( x)
Q , there is the
x a ( x)
the interval [a;b] and f(a)=(a)=0, then having limit lim
limit which equals
f ( x)
f ( x)
lim
x a ( x )
xa ( x) .
lim
In the following examples, we will use the following three-step process:
Step 1. Check that the limit of
0
f ( x)
is an indeterminate form of type . If it is
0
g ( x)
not, then L’Hospital’s Rule cannot be used.
Step 2. Differentiate f and g separately. [Note: Do not differentiate
the quotient rule!]
f ( x)
using
g ( x)
f ( x )
. If this limit is finite, , or , then it is equal
g ( x )
0
f ( x)
to the limit of
. If the limit is an indeterminate form of type , then simplify
0
g ( x)
f ( x )
algebraically and apply L’Hospital’s Rule again.
g ( x )
Step 3. Find the limit of
x2 4
2x
lim
2(2) 4
Example 6: lim
x 2 x 2
x 2 1
II. Indeterminate Form of the Type
We have previously studied limits with the indeterminate form
as shown in
the following examples:
3x 2
2
Example 7:
lim
3x 5 x 7
x 2 x
2
3x 1
x2
2
x 2 x
lim
x2
5
x
lim
x 2 3
x
3
5x
x2
3x
x2
7
x2
1
x2
7
300 3
x2
lim
1
x 2 0 0 2
x2
However, we could use another version of L’Hospital’s Rule.
L’Hospital’s Rule for Form
Suppose that f and g are differentiable functions on an open interval
containing x = a, except possibly at x = a, and that lim f ( x) and
x a
f ( x)
has a finite limit, or if this limit is or
lim g ( x) . If lim
x a g ( x)
xa
f ( x)
f ( x)
. Moreover, this statement is also true
lim
x a g ( x)
x a g ( x)
in the case of a limit as x a , x a , x , or as x .
, then lim
III. Indeterminate Form of the Type 0
Indeterminate forms of the type 0 can sometimes be evaluated by rewriting the
product as a quotient, and then applying L’Hospital’s Rule for the indeterminate
forms of type
0
or .
0
1
ln x
x2
x
Example 8: lim x ln x lim
lim
lim ( x) 0
lim
1
1
x
x 0
x 0
x 0
x
0
x 0
2
x
x
IV. Indeterminate Form of the Type
A limit problem that leads to one of the expressions
() () , () () , () () ,
() ()
is called an indeterminate form of type . Such limits are indeterminate
because the two terms exert conflicting influences on the expression; one pushes it
in the positive direction and the other pushes it in the negative direction. However,
limits problems that lead to one the expressions
() () , () () , () () , () ()
are not indeterminate, since the two terms work together (the first two produce a
limit of and the last two produce a limit of ). Indeterminate forms of the
type can sometimes be evaluated by combining the terms and manipulating
0
or .
0
1
1
cos x 1
sin x x
Example 9: lim
lim
lim
sin x x0 x sin x x0 x cos x sin x
x 0 x
sin x
0
0
lim
2
x 0 x sin x cos x cos x
the result to produce an indeterminate form of type
V. Indeterminate Forms of the Types 0 0 , 0 , 1
Limits of the form lim f ( x)g ( x) or lim f ( x)g ( x ) frequently give rise to
x
0
0
indeterminate forms of the types 0 , , 1 . These indeterminate forms can
x a
sometimes be evaluated as follows:
(1) y f ( x)g ( x )
(2) ln y ln f ( x)g ( x ) g ( x) ln f ( x)
(3) lim ln y lim g ( x) ln f ( x)
x a
x a
The limit on the righthand side of the equation will usually be an indeterminate
limit of the type 0 . Evaluate this limit using the technique previously described.
Assume that lim g ( x) ln f ( x)= L.
xa
(4) Finally, lim ln y L ln lim y L lim y e L .
x a
x a
x a
x
Example 10: Find lim x .
x 0
This is an indeterminate form of the type 0 0 . Let y x x ln y ln x x x ln x .
1
ln x
x
lim ln y lim x ln x lim 1 lim 1 lim x 0.
x 0
x 0
x 0
x 0
x 0
x
x2
Thus, lim x x e 0 1 .
x 0
Taylor’s formula
Suppose we’re working with a function f(x) that is continuous and has n+1
continuous derivatives on an interval about x = 0. We can approximate f near 0 by
a polynomial Pn (x) of degree n:
• For n = 0, the best constant approximation near 0 is P0 ( x) f (0) which matches
f at 0.
• For n = 1, the best linear approximation near 0 is P1 ( x) f (0) f (0) x . Note
that P1 matches f at 0 and P1 matches f at 0.
• For n = 2, the best quadratic approximation near 0 is
f (0) 2
Note that P2 , P2 , and P2 match f , f , and
P2 ( x) f (0) f (0) x
x .
2!
f , respectively, at 0.
Continuing this process,
Pn ( x) f (0) f (0) x
f (0) 2
f ( n ) (0) n
x ...
x .
2!
n!
This is the Taylor polynomial of degree n about 0 (also called the Maclaurin series
of degree n). More generally, if f has n+1 continuous derivatives at x = a, the
Taylor series of degree n about a is
n
k 0
f ( k ) (a)
f (a)
f ( n ) (a)
( x a) ( k ) f (a) f (a)( x a)
( x a) 2 ...
( x a) n .
k!
2!
n!
This formula approximates f (x) near a. Taylor’s Theorem gives bounds for the
error in this approximation:
Taylor’s Theorem:
Suppose f has n+1 continuous derivatives on an open interval containing a. Then
for each x in the interval,
n f ( k ) (a)
f ( x )
( x a) ( k ) Rn1 ( x) ,
k 0 k!
( n 1)
f
(c)
where the error term Rn1 ( x ) satisfies Rn1 ( x)
( x c) ( n1) for some c between
(n 1)!
a and x.
This form for the error Rn1 ( x ) , derived in 1797 by Joseph Lagrange, is called the
Lagrange formula for the remainder. The infinite Taylor series converges to f ,
f ( x)
if and only if lim Rn ( x) 0 .
k 0
f ( k ) (a)
( x a) ( k ) ,
k!
n
LECTURE 8
INVESTIGATION OF THE FUNCTION. EXTREMUM OF THE
FUNCTION. NECESSARY AND SUFFICIENT CONDITIONS FOR THE
EXISTENCE OF AN EXTREMUM. CONVEXITY, CONCAVITY AND
INFLECTION POINTS. ASYMPTOTE. THE OVERALL STUDY OF
DESIGN FEATURES
1.
2.
3.
4.
Monotonic conditions. Extremum of function
Convexity and concavity. Point of inflection
Аsymptotes
General Scheme for the Investigation of the Graph of a Function
MONOTONIC CONDITIONS. EXTREMUM OF FUNCTION
Theorem 1. Let f ( x ) be continuous on [a, b] and differentiable on (a, b). f ( x ) is a
constant
function if and only if f ' ( x ) 0 for all x ( a ,b ) .
Definition 1.
A function f ( x ) is said to be monotonic increasing (resp.
monotonic decreasing )
or simply increasing ( resp. decreasing ) on an interval I if and only if x1 , x2 I , if
x1 x2
then f ( x1 ) f ( x2 ) (resp. x1 , x2 I , if x1 x2 , then f ( x1 ) f ( x2 ) ).
Definition 2.
A function f ( x ) is said to be strictly increasing ( resp. strictly
decreasing ) on an
interval I if and only if x1 , x2 I , if x1 x2 then f ( x1 ) f ( x2 ) (resp. x1 , x2 I , if
x1 x2 ,
then f ( x1 ) f ( x2 ) ).
Theorem 2. Let f ( x ) be continuous on [a, b] and differentiable on (a, b). Then
(a) if f ' ( x) 0, x ( a ,b ), f ( x ) is strictly increasing on [a, b]; and
(b)if f ' ( x) 0, x ( a ,b ), f ( x ) is strictly decreasing on [a, b].
Definition 3. A neighborhood of a point x 0 is an open interval containing x 0 ,
i.e. (x 0 δ , x 0 δ ) is a neighborhood of x 0 for some δ 0 .
Definition 4.
A function f ( x ) is said to attain a relative maximum (minimum )
at a point x 0 if
f (x) f (x 0 ) ( f (x) f (x 0 ) ) in a certain neighborhood of x 0 , i.e. δ 0 such that
f (x) f (x 0 ) ( f (x) f (x 0 ) ) for x x 0 δ .
Theorem (Fermat Theorem).
Given f ( x ) is a point defined on (a, b ) and differentiable at a point x 0 if f ( x ) has an
extreme value ( max. or min ) x 0 , then f ' ( x 0 ) 0 .
f ' ( x 0 ) 0 f ( x ) has maximum or minimum at x 0 .
Note
Definition 5.
(a) A turning point is a maximum or minimum point.
(b) If f ' ( x) 0 , then x is called a critical or
stationary value and its corresponding point on the graph y f ( x) is called
stationary point.
Notes
1.
turning point
stationary point
turning point + differentiable stationary point
stationary point
turning point
2.
3.
CONVEXITY AND CONCAVITY. POINT OF INFLECTION
Let f x be a function differentiable on an interval J. The function f is called
convex (concave) on J, if all points of its graph on J lie above (below) any tangent
line to G f on this interval (excepting point of tangency). Let f be continuous at a
point x0 . If there exists 0 such that f is concave (convex) in N x0 and convex
(concave) in N x0 , the point x0 is called the point of inflection of f x .
The second Derivative Test for Concavity and Convexity:
If f ' ' x 0 , for each x J , then f x is convex on J,
if f ' ' x 0 , for each x J , then f x is concave on J.
It follows:
If f x is continuous at x0 and f ' ' x 0 ( f ' ' x 0 ) in N x0 and f ' ' x 0
( f ' ' x 0 ) in N x0 , then x0 is a point of inflection.
Moreover: If x0 is a point of inflection of f, then either f ' ' x0 0 or f ' ' x0
doesn’t exist.
If f is three times differentiable at a point x0 , f ' ' x0 0 and f ' ' ' x 0 , then x0 is
a point of inflection.
Definition 6.
Given that f (x) is continuous on [a, b] , if any x1 , x2 (a, b) such
that
x1 x 2 f ( x1 ) f ( x 2 )
2
2
(i) f
Concave Downward
(ii)
x x 2 f ( x1 ) f ( x 2 )
f 1
2
2
Concave Upward
Theorem 3. If f (x) is a function on [a, b] such that f (x) is second differentiable on
(a, b) then
(i) f ' ' ( x) 0 iff f (x) is concave upward on (a, b)
(ii) f ' ' ( x) 0 iff f (x) is concave downward on (a, b) .
Definition 7.
Let f ( x ) be a continuous function. A point (c, f (c)) on the graph
of f is a point of inflexion (point of inflection) if the graph on one side of this
point is concave downward and concave upward on the other side. That is, the
graph changes concavity at x c .
Note
A point of inflexion of a curve y f ( x) must be a continuous point
but need not be differentiable there. In Figure (c), R is a point of
inflexion of the curve but the function is not differentiable at x 0 .
Theorem 4. If f ( x ) is second differentiable function and attains a point of inflexion at
x c , then
f ' ' (c ) 0 .
Note:
(i)
max. or min. point but not derivative.
(ii) point of inflexion may not be obtained by solving f '' ( x) 0 where f ' (c)
and f ' (c) such that f ' (c)f ' (c) 0 .
(iii) Let f ( x ) be a function which is second differentiable in a neighborhood of a
point of inflexion iff f ' ( x ) does not change sign as x increases through (sign
gradient test)
– if f ' (c) 0 and f ' (c)f ' (c) 0 , then f ( x ) attains a relative max. or relative min.
– if f ' (c) 0 and f ' (c)f ' (c) 0 , then f ( x ) attains an inflexion point at c .
ASYMPTOTES
If a function f gets close to a certain number L when x gets larger and
larger, then we say that the limit as x goes to infinity is L and we write:
lim f ( x) L . Likewise, if f gets close to L when x gets smaller and smaller,
x
then the limit as x goes to negative infinity is L and we write: lim f ( x) L . In
x
both cases, the line y = L is a horizontal asymptote of f.
1
1
0 because when x is very large,
is close to 0. The x-axis is
x
x
x
y
a horizontal asymptote of the function f ( x) 1 :
x y 1
Example:
lim
x
x
If a function f ( x) gets larger and larger as x gets close to a number a, then it
“goes to infinity” and we write: lim f ( x) . The line x = a is a vertical
x a
asymptote of f. Similarly, if f ( x) gets smaller and smaller as x gets close to a,
then it “goes to negative infinity” and we write: lim f ( x) . Again, the line x
x a
= a is a vertical asymptote of f.
An important result:
If lim | f (x) | then lim 1 0 . This is because one over a very large positive
x a
xa
f ( x)
number and one over a huge negative number are both close to 0.
Inclined asymptotes have an equation y = kx + q and their position is arbitrary
except vertical. In order a straight line y = kx + q be an asymptote, the coefficients
k and q must satisfy at least one pair of the following conditions
f x
and b lim f x kx (k and b are numbers).
x x
x
k lim
Naturally these limits must be finite real numbers. A certain function can have
maximally two inclined asymptotes.
GENERAL SCHEME FOR THE INVESTIGATION
OF THE GRAPH OF A FUNCTION
The following information is useful for sketching the graph of y f (x)
(1)
The domain of f (x) , i.e. the range of values of x within which y is welldefined.
(2)
Determine whether f (x) is periodic, odd or even, so that the graph may
be symmetric about the coordinate axes or about the origin.
(3)
Turning points and monotonicity of f (x) .
(4)
Inflectional points and convexity of f (x) .
(5)
Asymptotes including horizontal, vertical and oblique ones (if any).
(6)
Some special points on the graph, such as intercepts.
LECTURE 9
ANTIDERIVATIVE. INDEFINITE INTEGRAL AND ITS PROPERTIES.
TABLE OF INTEGRALS. DIRECT INTEGRATION, INTEGRATION
WITH THE CHANGE OF VARIABLES AND BY PARTS
LECTURE PLAN:
1. Antiderivative and indefinite integral
2. Main methods of integration
ANTIDERIVATIVE AND INDEFINITE INTEGRAL
Concept of antiderivative and indefinite integral
Definition 1. A function F is an antiderivative of f on an interval I if
F x f x for all x in I .
Theorem 1. If F is an antiderivative of f on an interval I , then G is an
antiderivative of
f on the interval I if and only if G is of the form G x F x C , for all x in I
where C is a constant.
The constant C is called the constant of integration.
The family of functions represented by G is the general antiderivative of f
G x F x C is the general solution of the differential equation G x F x .
Notation for antiderivatives
dy
f x , we solve for dy ,
When solving a differential equation of the form
dx
giving us the equivalent differential form dy f x dx . The operation of finding
all solutions of this equation is called antidifferentiation or indefinite integration
and is denoted by an integral sign . The general solution is denoted by
Variable of
Integration
y f x dx F x
Integrand
C
Constant of
Integration
Definition. Suppose f is a function defined on an interval I and suppose further
that f has an antiderivative on the interval I. The family of all antiderivatives of f is
called the indefinite integral of f and is denoted by the symbol f x dx (read the
integral of f with respect to x). In this notation the function is called the integrand
of the indefinite integral. The process of finding the indefinite integral is called
integration or, sometimes anti-differentiation. More specifically, given a function
f, the expression “integrating f” means “finding the indefinite integral of f”.
There is a difference between an antiderivative and the indefinite integral. An
antiderivative is a member of the family of functions represented by the indefinite
integral. In the above example, 2xdx x 2 C , where C is a constant. The
functions x 2 , x 2 1and x 2 3 are all members of the family 2xdx and are all
antiderivatives of the function 2x . It is also correct to write 2 xdx x 2 3 K ,
where K is a constant. The difference is the choice of constants, where in this
case C K 3 .
Notice that all these curves are “parallel” in the sense that they never cross
each other since they are translates of the function x 2 . To get a particular
antiderivative one needs to know a specific point the antiderivative passes through.
For example, if we want the curve that is a member of the family 2xdx that passes
through
the
point 1,3 ,
have f x x2 C and f 1 3 .
we
So 3 f 1 1 C C 2 and it follows that f x x2 2 .
The process of integration involves finding one antiderivative of the given
function so that the indefinite integral is that one antiderivative plus the constant of
integration. This can be easy for certain functions because we know the
2
differentiation
formulae.
For
example,
because
d
sin x cos x ,
dx
it
follows
that cos xdx sin x C , C a constant. Here are the integration formulae that follow
directly from the appropriate differentiation formulae: The symbol C is a constant
(called the constant of integration). For any differentiable function f it is always
the case that f x f x dx meaning that the function f is a member of the
family f x dx .
1.
0dx C
2.
d x r 1
x r 1
x r x r dx
C for any constant r 1 . Notice that when r 0 ,
dx r 1
r 1
the
formula reduces to 1dx x C or just dx x C .
3.
4.
5.
6.
7.
8.
d
1
dx
ln x x 1dx
ln x C
dx
x
x
d x
e e x e x dx e x C
dx
d
sin x cos x cos xdx sin x C
dx
d
cos x sin x sin xdx cos x C
dx
d
tan x sec2 x sec2 xdx tan x C
dx
d
cot x csc2 x csc2 xdx cot x C
dx
d
sec x sec x tan x sec x tan xdx sec x C
dx
d
10. csc x csc x cot x csc x cot xdx csc x C
dx
9.
In fact every differentiation formula provides an integration formula. Consider the
differentiation formula
d
d
f x .
cf x c
dx
dx
This is a statement that two
derivatives are equal. So the families of antiderivatives are also the same families.
Therefore cf x dx c f x dx . In the same manner,
d
d
d
f x g x f x g x dx f x dx g x dx .
f x g x
dx
dx
dx
For reference, number these formulae as
11. cf x dx c f x dx
12. f x g x dx f x dx g x dx
These two formulae and those above can be combined to produce integrals of more
complicated functions as shown in the examples below.
Properties of indefinite integral. The table of elementary integrals
1. dF x F x C
2. d f x dx f x dx
3. f x dx f x dx, где 0
f x g x dx f x dx g x dx
5. If f x dx F x C , then f ax b dx 1 F ax b C , where a 0 .
a
4.
The table of indefinite integrals
1x n dx
x n 1
C , n 1
n 1
( x )
n
dx
dx
1 ( x ) n 1
C,
n 1
1
n 1
2dx ln x C
x
3
ax
a
dx
C
ln a
x
x
4e dx e C
a
x
e
x
5sin xdx cos x C
6cos xdx sin x C
sin( x )dx cos( x ) C
7 dx tgx C
2
cos
8 dx
2 ctgx C
sin
x
x
cos
x
sin x
ln x C
1 ax
dx
C
ln a
dx
1
e x C
1
1
cos( x )dx sin( x ) C
2
dx
2
( x )
1
tg ( x ) C
dx
1
ctg ( x ) C
( x )
9
dx
a x
1 dx
x2 a2
1 dx
x2 a2
2
dx
1 x
2
arcsin
x
C
a
a ( x )
dx
1
x
arctg C
a
a
1 xa
ln
C
a xa
ln x
dx
2
( x )
a2
2
dx
( x )
x2 C
2
2
2
a2
dx
x
1
2
1
arcsin
x
a
C
x
1 1
arctg
C
a
a
1 1 x a
ln
C
x a
a
ln x
x 2
C
Examples
Example 1: Evaluate x 4 4 x 5 dx
Solution:
x
4
4 x 5 dx x 4 dx 4 xdx 5dx (Formula 12, applied twice)
x 4 dx 4 xdx 5 dx
x2
x5
4 5 x C
5
2
1
x5 2 x 2 5 x C
5
(Formula 11)
(Formula 2)
(Arithmetic)
Example 2: Evaluate 6 t 6 t dt
Solution:
6
t 6 t dt 6 tdt 6 tdt
6 tdt 6 tdt
1
6 t 2 dt t 6 dt
1
t 32 t 7 6
6
C
3 7
2
6
7
3
6
4t 2 t 6 C
7
(Formula 12)
(Formula 11)
(Algebra)
(Formula 2)
(Arithmetic)
Example 3: Find all the antiderivatives of x3 5sec2 x .
Solution: This question is just another way to ask that x3 5sec2 x dx be
evaluated.
3
2
3
2
(Formula 12)
x 5sec x dx x dx 5sec xdx
x3 dx 5 sec 2 xdx
x4
5 tan x C
4
1
x 4 5 tan x C
4
Example 4: Evaluate
sin
d
cos 2
(Formula 11)
(Formulae 2 and 7)
(Arithmetic)
Solution:
sin
cos
2
1 sin
d
d
cos cos
sec tan d
(Algebra)
(Trig Relationships)
(Formula 9)
sec C
Example 5: Evaluate cos u 1 tan u du .
Solution: cos u 1 2 tan u du cos u 2sin u du (Algebra)
cos ud 2sin udu
(Formula 12)
cos ud 2 sin udu
(Formula 11)
sin u 2 cos u C
sin u 2cos u C
(Formulae 5 and 6)
(Algebra)
Example 6: Suppose f x x3 2x 3cos x . Find the function f
Solution: f x f x dx x3 2 x 3cos x dx
x3dx 2 xdx 3cos xdx
(Formula 12)
x3dx 2 xdx 3 cos xdx
(Formula 11)
x2
x4
2 3sin x C1
4
2
1
x 4 x 2 3sin x C1
4
1
f x f x dx x 4 x 2 3sin x C1 dx
4
1
x 4 dx x 2 dx 3sin xdx C1dx
4
1
x 4 dx x 2 dx 3 sin xdx C1 dx
4
1 x5 x3
3 cos x C1 x C2
4 5 3
1 5 1 3
x x 3cos x C1 x C2
20
3
(Formulae 2 and 6)
(Algebra)
(Formula 12)
(Formula 11)
(Formulae 2 and 5)
(Algebra)
Example 7: Suppose a particle is moving along a line so that the velocity at
1
t
time t is given by v t sin t . The displacement s t at time t is zero. Find
the displacement at any time t.
Solution: s t v t dt sin t dt
1
t
ln t cos t C
dt
sin tdt
t
(Formula 12)
(Formulae 3 and 5)
(Algebra)
ln t cos t C
Since s 0 , 0 s ln cos C ln 1 C C 1 ln .
So s t ln t cos t 1 ln .
MAIN METHODS OF INTEGRATION
Integration by substitution ( or change of variable )
Steps for Integrating by Substitution—Indefinite Integrals:
1.
Choose a substitution u = g(x), such as the inner part of a composite
function.
2.
Compute du g ( x)dx .
3.
Re-write the integral in terms of u and du.
4.
Find the resulting integral in terms of u.
5.
Substitute g(x) back in for u, yielding a function in terms of x only.
6.
Check by differentiating.
If f(x) is continous function, F(x)- its antiderivative and φ(х)- differentiable
function, then
x t
f x x dx x dx dt f t dt F t C F x C
In the particular case
Example 8. To find
x
x t
x dx x dx dt
dt
ln t C ln x C
t
3x 2
4 x3 dx . Notice that the numerator is the derivative of the
denominator
Let y 4 x3 . Differentiating gives
dy
3 x 2 and hence dy 3x 2 dx .
dx
Substituting this change of variable the integral becomes
1
y dy ln y C
Now by expressing this result in terms of x we have shown that
3x 2
3
4 x3 dx ln(4 x ) C .
Integration by parts
By the Product Rule for Derivatives,
d
f ( x) g ( x) f ( x) g ( x) g ( x) f ( x) . Thus,
dx
f (x)g (x) g(x) f (x) dx f (x)g(x) f (x)g (x) dx g(x) f (x) dx
f ( x) g ( x) f ( x) g ( x) dx f ( x) g ( x) g ( x) f ( x) dx . This formula for integration
by parts often makes it possible to reduce a complicated integral involving a
product to
a simpler integral. By letting u f ( x) du f ( x)dx
dv g ( x)dx v g ( x)
we get the more common formula for integration by parts:
udv uv vdu .
Example 9. Find
x ln xdx .
1
x
1
2
1
x ln xdx (ln x)( x dx) udv uv vdu (ln x) x 2
2
1
1
11
1 2 1 1 2
x dx x 2 ln x x 2 C
x dx x ln x
2
2
22
2 x 2
1 2
1
x ln x x 2 C .
2
4
Let u ln x and dv xdx du dx and v x dx x 2 . Thus,
It is possible that when you set up an integral using integration by parts, the
resulting
integral will be more complicated than the original integral. In this case, change
your
substitutions for u and dv.
LECTURE 10
INTEGRATION OF SIMPLE RATIONAL FRACTIONS. INTEGRATION
OF RATIONAL FRACTIONS
LECTURE PLAN:
1.
2.
3.
Selecting the proper rational fraction
Integrating Proper Rational Functions
Integrating Improper Rational Functions
Selecting the proper rational fraction
Suppose f ( x)
P( x)
Q( x )
is a rational function; that is, P (x ) and Q (x) are
polynomial functions. If the degree of P (x ) is greater than or equal to the degree
of Q (x) , then by long division, f ( x)
R( x)
P( x)
R( x)
where
is a proper
S ( x)
Q ( x)
Q( x)
Q( x)
rational fraction; that is, the degree of R (x ) is less than the degree of Q (x) . A
theorem in advanced algebra states that every proper rational function can be
expressed as a sum
R ( x)
F1 ( x) F2 ( x) Fn ( x)
Q( x)
where F1 ( x), F2 ( x), ..., Fn ( x) are rational functions of the form
Ax B
A
or
2
k
(ax bx c) k
(ax b)
in which the denominators are factors of Q (x) . The sum is called the partial
R( x)
. The first step is finding the form of the partial
Q ( x)
R( x)
fraction decomposition of
is to factor Q (x) completely into linear and
Q ( x)
fraction decomposition of
irreducible quadratic factors, and then collect all repeated factors so that Q (x) is
expressed as a product of distinct factors of the form
(ax b) m and (ax 2 bx c) m .
From these factors we can determine the form of the partial fraction decomposition
using the following two rules:
Linear Factor Rule: For each factor of the form (ax b) m , the partial fraction
decomposition contains the following sum of m partial fractions:
Am
A1
A2
2
ax b (ax b)
(ax b) m
where A1, A2, . . ., Am are constants to be determined.
Quadratic Factor Rule: For each factor of the form (ax 2 bx c) m , the partial
fraction decomposition contains the following sum of m partial fractions:
Am x Bm
A1 x B1
A2 x B2
2
2
2
ax bx c (ax bx c)
(ax 2 bx c) m
where A1, A2, . . ., Am, B1, B2, …, Bm are constants to be determined.
I. Integrating Proper Rational Functions
3x 17
dx .
x 2x 3
x 2 2 x 3 ( x 3)( x 1) using the Linear Factor Rule, we get
3x 17
3x 17
A
B
3x 17 A( x 1) B( x 3) after
2
x 2 x 3 ( x 3)( x 1) x 3 x 1
multiplying by ( x 3)( x 1) . If we let x 3 , then 8 4 A A 2 ; if we let
3x 17
dx =
x 1 , then 20 4B B 5 . Thus,
2
x 2x 3
3x 17
1
1
dx 2
dx 5
dx 2 ln x 3 5 ln x 1 C .
( x 3)( x 1)
x 3
x 1
3x 4
Example 11: Find
dx .
2
x 4x 4
x 2 4 x 4 ( x 2)( x 2) ( x 2) 2 by the Linear Factor Rule, we get
B
3x 4
3x 4
A
3 x 4 A( x 2) B after multiplying
2
2
x 2 ( x 2) 2
x 4 x 4 ( x 2)
Example 10: Find
2
by ( x 2) 2 . If we let x = 2, then 2 B ; if we let x = 3, then
5 A B A 2
A 3 . Thus,
3 ln x 2
3x 4
dx 3
x 4x 4
2
1
dx 2
x2
2
C.
x2
II. Integrating Improper Rational Functions
1
dx
( x 2) 2
Although the method of partial fractions only applies to proper rational
functions, an improper rational function can be integrated by performing long
P( x)
is a rational function where P (x ) and
Q( x )
Q (x) are polynomial functions and the degree of P (x ) is greater than or equal to the
R( x)
P( x)
R( x)
degree of Q (x) , then by long division, f ( x)
where
is a
S ( x)
Q ( x)
Q( x)
Q( x)
R( x)
proper rational fraction. Since
is a proper rational function, it can be
Q ( x)
division (or synthetic division). If f ( x)
decomposed into partial fractions.
Example 12: Find
x3 1
dx .
x 1
1 0 0 1
By synthetic division, 1
x3 1
2
.
x2 x 1
x 1
x 1
1 1 1
1 1 1 2
x3 1
dx
Thus,
x 1
ln x 1 C .
( x 2 x 1) dx 2
1
1
1
dx x 3 x 2 x
x 1
3
2
LECTURE 11
INTEGRATION OF EXPRESSIONS CONTAINING TRIGONOMETRIC
FUNCTIONS. INTEGRATION OF IRRATIONAL FUNCTIONS
INTEGRATION OF IRRATIONAL FUNCTIONS
Type 1.
ax b
.
cx d
ax b
Use the substitution u n
.
cx d
Integrand contains
n
Type 2.
Px Q
dx .
ax2 bx c
d
Write Px Q as Px Q p (ax2 bx c) q .
dx
Type 3.
Integrand contains
,
or
This was discussed in "trigonometric substitutions above". Here is a summary:
1. For
, use
.
Integral is of the form
2. For
3. For
, use
, use
.
.
Type 4.
1
Integral is of the form
dx .
( px q) ax2 bx c
1
Use the substitution u
.
px q
Type 5.
Other rational expressions with the irrational function
1. If
, we can use
.
2. If
3. If
, we can use
can be factored as
a( x )
use u
.
x
4. If
use
and
.
, we can
can be factored as
Frequently occuring integrals of irrational functions are:
, we can
dx
2
ax bx c
, where
2
a 0 and ax bx c is possitive on an interval. We can exclude the case, that the
polynomial ax 2 bx c has a double root.
1
1
Taking the factor
(if a 0 ) or
(if a 0 ) we reduce the integral to the form
a
a
dx
dx
or
x 2 px q
x 2 px q
, leading (by means of substitution) to the
integrals:
dx
x2 k 2
ln x x 2 k 2 C or
dx
k 2 x2
arcsin
x
C , respectively.
k
3. INTEGRATION OF TRIGONOMETRIC FUNCTIONS
1. Given an integral
Rsin x, cos xdx , i.e. the integrand is a rational function in
x
the integral is reduced to an
2
x
2dt
2t
integral of a rational function. If t tan , then x 2arctan t , dx
,
sin
x
2
1 t 2
1 t 2
terms of sin x and cos x . By the substitution t tan
and cos x
1 t 2
.
1 t 2
2. If R sin x, cos x = Rsin x, cos x , then tgx t .
3. If R sin x, cos x = - Rsin x, cos x , then cos x t .
If Rsin x, cos x =- Rsin x, cos x , then sin x t .
4. sin m x cos n xdx , т and п – even non-negative integers, then
sin 2 x
1 cos 2 x
2
; cos 2 x
1 cos 2 x
2
,
sin x cos x
cos nx cos mxdx;
5. For integrals
following formulas:
1
sin 2 x .
2
sin nx sin mxdx; sin nx cos mxdx
we use
1
cos cos ;
2
1
sin sin cos cos ;
2
1
sin cos sin sin
2
cos cos
6. tg n xdx,
ctg
n
xdx , then tg 2 x
1
1
cos 2 x
and
ctg 2 x
1
1.
sin 2 x
LECTURE 12
THE DEFINITE INTEGRAL. PROBLEMS LEADING TO THE DEFINITE
INTEGRAL. THE NEWTON-LEIBNIZ FORMULA.
LECTURE PLAN:
1. Concept of definite integral
2. Main properties of definite integrals
3. Applications of definite integrals
CONCEPT OF DEFINITE INTEGRAL
Definition 1.
Let f (x) be a continuous function defined on [a, b] divide
the interval by the points a x0 x1 x 2 x n 1 b from a to b into n
subintervals. (not necessarily equal width) such that when n , the length of
each subinterval will tend to zero.
( xi xi 1 ) f (i ) exists
In the ith subinterval choose i [ xi 1 , xi ] for i 1,2, , n . If nlim
and is independent of the particular choice of x i and i , then we have
b
a
n
f ( x) dx lim
n
(x x
i 1
i
i 1
) f (i ) .
Remark. For equal width, i.e. divide [a, b] into n equal subintervals of length, i.e.
n
n
b
ba
ba
f
(
)
h
lim
f ( i )
, we have a f ( x) dx nlim
.
i
n
n
n
i 1
i 1
Choose i xi and xi a ih
h
b
a
f ( x) dx lim
n
n
f (a
i 1
ba
i ) h or
n
b
a
f ( x) dx lim
n
n 1
f (a
i 0
ba
i) h .
n
MAIN PROPERTIES OF DEFINITE INTEGRALS
Properties of Definite Integrals
P1. The value of the definite integral of a given function is a real number,
depending on its lower and upper limits only, and is independent of the choice of
the variable of integration, i.e.
b
b
b
f
(
x
)
dx
f
(
y
)
dy
a
a
a f (t )dt .
P2.
b
a
a
P3.
a
a
f ( x)dx f ( x)dx
b
b
b
f ( x)dx f ( x)dx 0 dx 0
b
a
Let a c b , then
P4.
b
a
c
b
a
c
f ( x)dx f ( x)dx f ( x)dx .
P5. Comparison of two integrals
b
b
If f ( x) g ( x) x (a, b) , then a f ( x)dx a g ( x)dx
P6. Rules of Integration
If f ( x), g ( x) are continuous function on [a, b] then
b
b
(a) a kf ( x)dx k a f ( x)dx for some constant k.
(b) a f ( x) g ( x)dx a f ( x)dx a g ( x)dx .
b
b
b
Newton – Leibniz formula
Comparing the two formulas of the curvilinear trapezoid area, we make the
conclusion: if F ( x ) is primitive for the function f ( x ) on a segment [ a, b ] ,
then
b
a
b
f x dx F x F b F a
a
This is the famous Newton – Leibniz formula. It is valid for any function f ( x ),
which is continuous on a segment [ a , b ] .
(i) If f ( x) f ( x) (Even Function)
then
a
a
(ii)
a
f ( x) dx 2 f ( x) dx .
0
If f ( x) f ( x) (Odd Function)
Then
a
a
f ( x) dx 0 .
Integration by substitution
If the function u = g(x) has a continuous derivative on [a, b] and f is
continuous on the range of g, then
g (b )
b
f ( g ( x))dx
a
f (u )du
g (a)
Steps for Integrating by Substitution—Definite Integrals
1.
Choose a substitution u = g(x), such as the inner part of a composite
function.
2.
Compute du g ( x)dx . Compute new u-limits of integration g(a) and g(b).
3.
Re-write the integral in terms of u and du, with the u-limits of integration.
4.
Find the resulting integral in terms of u.
5.
Evaluate using the u-limits. No need to switch back to x’s!
Integration by Parts
Let u and v be two functions in x . If u ' ( x) and v' ( x) are continuous on a, b ,
then
uv' dx uv vu' dx or
b
b
b
a
a
a
udv uv vdu .
b
b
b
a
a
a
LECTURE 14
APPLICATIONS TO THE COMPUTATION OF THE INTEGRALS OF
PLANE FIGURES AREAS. CALCULATION THE ARC LENGTH, THE
AMOUNT OF BODY ROTATION. THE IMPROPER INTEGRAL
LECTURE PLAN:
1. Application to the computation of definite integrals of plane figures areas
2. Calculations of the length, the amount of body rotation
3. The improper integral
APPLICATIONS OF DEFINITE INTEGRALS
The areas of plane figures
If a continuous curve is defined in rectangular coordinates by the equation
y f ( x) ( f ( x) 0) the area of the curvilinear trapezoid bounded by this curve, by
two vertical lines at the
points x=a and x =b and by a segment of the x-axis a x b , is given by the
formula
b
S f ( x) dx .
a
In the more general case, if the area S is bounded by two continuous curves
y f1 ( x) and y f 2 ( x) and by two vertical lines x=a and x=b, where
f1 ( x) f 2 ( x) when a x b , we will then have:
b
S f 2 ( x) f1 ( x)dx .
a
If the curve is defined by equations in parametric form x (t ) and y (t )
then the area of the curvilinear trapezoid bounded by this curve, by two vertical
lines (x=a and x=b), and by a segment of the x-axis is expressed by the integral
t2
S (t ) (t )dx ,
t1
where t1 and t 2 are determined from the equations a (t1 ) and
b (t 2 ) ( (t ) 0) on the interval t1 ,t 2 .
If a curve is defined in polar coordinates by the equation r f ( ) , then the
area of the sector AOB (Fig. 2), bounded by an arc of the curve, and by two radius
vectors OA and OB,
Fig. 2.
which correspond to the values 1 and 2 is expressed by the integral
S f ( ) d .
2
The arc length of a curve
The arc length s of a curve y=f(x) contained between two points with
abscissas x=a and x=b is
b
S 1 y2 dx .
a
If a curve is represented by equations in parametric form x (t ) and
y (t ) then the arc length s of the curve is
t2
S x2 y2 dt ,
t1
where t1 and t 2 are values of the parameter that correspond to the extremities of the
arc.
If a curve is defined by the equation r f ( ) in polar coordinates, then the
arc length s is
S r 2 r 2 d ,
where and are the values of the polar angle at the extreme points of the arc.
The volume of a solid of revolution
The volumes of solids formed by the revolution of a curvilinear trapezoid
[bounded by the curve y=f(x), the x-axis and two vertical lines x=a and x=b]
about the x-axis and y-axes are expressed, respectively, by the formulas:
b
b
Vx y dx and Vy 2 xydx .
2
a
a
The Area of a Surface of Revolution
The area of a surface formed by the rotation, about the x-axis, of an arc of the
curve y=f(x) between the points x=a and x=b, is expressed by the formula
b
b
ds
S x 2 y dx 2 y 1 y2 dx .
dx
a
a
x xt ,
If a curve is represented by equations in parametric form
y y t ,
then
t2
S 2 yt xt y 2 dt ,
2
2
t1
where t1 and t 2 are values of the parameter t.
LECTURE 15
COMPLEX NUMBER. COMPLEX NUMBERS IN TRIGONOMETRIC
AND EXPONENTIAL FORM
LECTURE PLAN:
1. Complex numbers
2. Complex numbers in trigonometric and exponential form
Complex numbers
The main theorem of algebra. An algebraic equation of the nth degree
xn+an-1xn-1+…+a1x+a0=0
has n zeros. (Without a proof.)
Example.
x3-1=0, (x-1)(x2+x+1)=0, x1=1,
x2+x+1=0; D=1-4=-3,
x 2, 3
1 3
- not real numbers.
2
1 is called the imaginary unit and is denoted by
Definition. The number
1 i
i 2 1 1 1 ;
i3=-i;
i4=i2.i2=(-1).(-1)=1; i5=i;
i6=-i and so on. i.e. is a multiple of 4.
Example. Find
i35= i32+3 =i3=-i.
Definition. A number of the form =a+bi is called a complex number,
where, i is the imaginary unit and а, b are real numbers. Numbers а and bi are
called the real and the imaginary parts of a complex number.
The following is a geometric interpretation of a complex number
у
М(a,b)
=a+bi
b
0
x
a
Each complex number corresponds to a point of the coordinate plane хOу. Each
real number is also a complex number with zero imaginary part, i.e. b=0. For
example, 32=32+0i.
Operations on complex numbers
1) Addition of complex numbers.
Consider complex numbers 1 =a1 +b1i, 2 =a2 +b2i.
Add them and combine the like terms
1 +2 =a1 +b1i +a2 +b2i= (a1 +a2) +i(b1+b2),
We add separately the real and the imaginary parts.
2) Subtraction 1–2 = (a1–a2) +i(b1–b2).
Geometrically, addition and subtraction of complex numbers is performed
exactly in the same way as addition and subtraction of the vectors with
corresponding coordinates.
y
1+2
b2
2
1-2
b1
0
1
a1 a2
x
3) Multiplication.
1 . 2 =(a1 +b1i)(a2 +b2i)= a1a2+b2ia1+b1a2i+b1b2i2=
=(a1a2-b1b2)+i( a1b2 +b1a2), т.к. i2=-1.
Multiplication of complex numbers is performed as usual multiplication of
algebraic expressions; after the multiplication the real and imaginary parts are
grouped separately.
The numbers =a +bi, =a –bi are called complex conjugates or simply
conjugates.
The product of two conjugates is a nonnegative real number a2+bia-abii2b2=a2+b2.
4) Division. Definition. The quotient
1
of two complex numbers is a
2
complex number such that 2 1 .
Suppose,
с di , then by the definition
2 (a2 b2i)(c di) a1 b1i ;
a2 c b2 d i(b2 c a2 d ) a1 b1i .
Due to equality of real and imaginary parts,
a2c b2d a1 ; b2c a2d b1 .
Solve the resulting system of linear equations
a2 c b2 d a1 a2 b2
b2 c a2 d b1 b2 a2
,
and find the real numbers c and d
(a2 b2 )c a1a2 b1b2 ,
a b a1b2
a a bb
c 1 22 1 2 2 , d 2 21
2 .
a2 b2
a 2 b2
2
2
The denominator of both expressions is the product of two conjugates,
so let us multiply both the numerator and the denominator of
1
by the
2
conjugate of the denominator
(a1 b1i )( a 2 b2 i ) (a1 a 2 b1b2 ) (b1 a 2 a1b2 )i
2
2
(a 2 b2 i )( a 2 b2 i )
a 2 b2
a1 a 2 b1b2
a 2 b2
2
2
(b1 a 2 a1b2 )i
a 2 b2
2
2
- as expected, the result equals .
Rule. In order to find the quotient of two complex numbers, it is enough to
multiply both the numerator and the denominator by the conjugate of the
denominator, and then separate the real and imaginary parts.
Examples. 1. Divide
(2 3i ) (2 3i )(5 6i ) 10 12i 15i 18 8 27i
8 27i
.
(5 6i ) (5 6i )(5 6i )
25 36
61
61 61
2. Solve the equation х2-2х+10=0 and check Viete's theorem for the complex
zeros
x1, 2 1 9 1 3i ;
x1 1 3i, x2 1 3i ;
x1+x2=2;
x1 .x2=12+32=10.
The magnitude and argument of a complex number. Plot the complex number
=a+bi on the xOy plane.
у
0
b
a
x
It is easy to see that 2=a2+b2 and tan
b
; the numbers ρ and are called the
a
absolute value (or magnitude) and the argument of a complex number respectively.
Trigonometric form of a complex number, de Moivre's formula
Use the above figure to express the real and imaginary parts a,b
of the complex number in terms of the magnitude and argument ρ, .
a=.cos, b=.sin.
Hence
=(cos+isin)
This is the trigonometric form of a complex number.
Each complex number has a unique trigonometric form because it has a
unique magnitude and argument.
Example. α= 2 2i .
2 2 4 2;
2
arctan
Therefore,
2 2i 2(cos
4
i sin
2
4
arctan 1
4
.
).
1. Multiplication. Consider complex numbers
1 =1 (cos1 +isin1) and 2 =2 (cos2 +isin2).
The product equals
1·2 = 1 (cos1 +isin1). 2 (cos2 +isin2)=
=1·2 [(cos1cos2 -sin1sin2)+i(sin1cos2 + cos1sin2);
1·2 = 1·2 [cos(1+2 )+isin(1+2 )].
Therefore, in order to multiply two complex numbers, it is necessary to multiply
their magnitudes and add the arguments.
2. Division. Multiply both the numerator and denominator by the conjugate
of the denominator
1 1 (cos 1 i sin 1 ) (cos 2 i sin 2 )
2 2 (cos 2 i sin 2 ) (cos 2 i sin 2 )
[cos 1 cos 2 sin 1 sin 2 i (sin 1 cos 2 cos 1 sin 2 )
1
2 (cos2 2 sin 2 2 )
1 [cos(1 2 ) i sin(1 2 )].
2
Therefore, in order to divide two complex numbers, it is necessary to divide their
magnitudes and subtract the arguments.
A power of a complex number. If 1=2=3=…=n=, then
n = · ··…·[cos(+ ++…+)+i(sin(+ ++…+)]= =n[cosn+isinn]
or
k =k[cosk+isink] - de Moivre's formula.
Roots of a complex number. Write down de Moivre's formula for k
1
n
using the fact the sine and cosine are periodic functions with the period T=2:
1
n
n cos
2k
n
i sin
2k
n
,
where k may be any of n integers 0,1,2,…,n-1.
Example. Solve the equation x3+1=0.
1-st method: (x+1)(x -x+1)=0 , x1=-1, x2,3
2
1
1
1
3
1
i;
2
4
2 2
3
2-nd method: x3=-1, x 1 , -1=cos+isin, which is a trigonometric form of a
number.
3
cos i sin 3 1(cos
k=0, x1 cos
3
i sin
3
k=1,
k=2, x3
2k
3
i sin
2k
3
) . Using the formula,
1
3
i
,
2
2
x2 cos i sin 1 ,
5
5
i sin
cos(2 ) i sin( 2 )
3
3
3
3
1
3
= cos( ) i sin( ) i
.
3
3
2
2
cos
ELABORATIONS OF PRACTICAL CLASSES
PRACTICAL CLASS № 1-2.
MATRIX. OPERATIONS WITH MATRICES. THE INVERSE MATRIX.
DETERMINANTS AND THEIR PROPERTIES. SYSTEMS OF LINEAR
EQUATIONS
Theoretical questions:
1. The concept of matrix;
2. Matrix operations: addition of matrices, multiplication of a matrix by a number,
matrix multiplication;
3. Calculating determinants of 2×2 and 3×3 matrices;
4. Solving systems of linear equations matrix method;
5. Solving systems of linear equations on Cramer's rule;
6. Investigation compatible system;
7. Solving systems of linear equations Gauss's method.
Classroom assignments:
1. Matrixes A 2 1 1 and
2 1 0
B
1 1 4
3 2 2
1 3
1 7
2. Find matrix Х, if: 2 2 4 X 2 8
0 5
3 9
3
2
3
4
3. Calculate: а)
;
5 4 2 5
б) АВ, if: A 2 0 , B 2 .
1 1
2
4. Calculate АВ and ВА, if:
5. Calculate D= ABC-3E, if:
are given. Find 3А+2В.
3
1
A 4 0 2 3 1, B 1
5
2
1 2 3
1
A 1 0 2 B 2 C 2 0 5
4 5 3
1
6. Find the values of matrix polynomials:
а) f ( x) 2x 2 5x 9 , если A 1 2
3 0
б)
в)
1
0
3
2
f ( x ) 3 x x 2,
f ( x) 3x
2
5 x 2,
3
5
A
1
A 0
2
2
0
1
4
2 1
7. Calculate determinants:
1
2 3
2
3 1
3
1 3 5 ;
1
5 4;
4 1 0 ;
0 3 1; 4 5 6
0
1 7 8
0
0 0
5 1
1
0
2
7
1 2
5 1 2 3
4 7 8 9
8. Prove that AB BA A B , if
2 1 1
1 4 1
A 5 1 6 ; B 2 5 0
0
3
3
3
9
2
9. Solve systems of linear equations matrix method by matrix method, by Cramer’s
formula, investigate systems on compatibility, and solve them by Gauss's method:
a.
x1 2 x 2 3x3 6
4 x1 5 x2 6 x3 9
7 x 8 x 6
2
1
b.
x1 2 x2 3x3 6
4 x1 5 x2 6 x3 15
7 x 8 x 9 x 24
2
3
1
c.
x1 2 x2 3x3 3
2 x1 6 x2 4 x3 6
3x 10 x 8 x 21
2
3
1
10. Investigate systems on compatibility and, in case of compatibility, solve them
by Gauss's method:
a.
x1 x 2 x 3 4
x1 2 x 2 3 x 3 0
2 x
3 x 3 16
1
b.
x1 2 x2 2 x3 3 x4 1
6 x 3 x 3 x x 9
1
2
3
4
7
x
x
x
2
x
1
2
3
4 8
3x1 9 x2 9 x3 10 x4 12
c.
x1 2 x 2 4 x3 3 x 4 0
3 x 5 x 6 x 4 x 0
1
2
3
4
4
x
5
x
2
x
3
x
1
2
3
4 0
3 x1 8 x 2 24 x3 19 x 4 0
Homework.
Theoretical material: The simplest problem of analytic geometry. Equations of a
straight line on a plane. Vectors.
Solve problems:
2 0
1 3
B 1 3 C
0 4
0 5
1.
Calculate D AB T C 2 , if: A 3
2.
Find АВС, if: A 4 3 B 28 93 C 7 3
7 5
38 126
2 1
3.
4 2
1 0 5
Prove, that AB BA A B , if:
2 3 1
1 1 0
A 2 3 5 B 1 2 3
3 1 2
5 0 1
4.
Solve systems of linear equations matrix method by matrix method and by
Cramer’s formula:
a.
2 x1 3x 2 x3 7
x1 2 x2 3x3 14
x x 5 x 18
2
3
1
b.
x1 x2 3x3 4
2 x1 6 x2 4 x3 6
3x 10 x 8 x 8
2
3
1
5. Investigate systems on compatibility and, in case of compatibility, solve them by
Gauss's method:
6 x1 5 x 2 4 x3 7 x 4 28
5 x 8 x 5 x 8 x 36
1
2
3
4
9 x1 8 x 2 5 x3 10 x 4 42
3 x1 2 x 2 2 x3 2 x 4 2
PRACTICAL CLASS № 3-5.
THE SIMPLEST PROBLEM OF ANALYTIC GEOMETRY. EQUATIONS
OF A STRAIGHT LINE ON A PLANE. ANALYTIC GEOMETRY IN
SPACE. VECTORS. SIMPLE OPERATIONS WITH VECTORS. THE
SCALAR, VECTOR AND MIXED PRODUCT OF VECTORS
Theoretical questions: 1. How Rectangular Cartesian coordinates are given on the
plane?
2. What names have Rectangular Cartesian coordinates axes?
3. What is connection between polar and Rectangular coordinates of points?
4. What formula find a distance between two points?
5. Dividing Formulas of segment in given ratio.
6. Equation of Straight (right) line with angular coefficients.
7. Common equation of Straight (right) line
8. Equation of Straight (right) line with given angular coefficients and passing
through a given point.
9. Equation of Straight (right) line passing through given two points.
10. Equation of Straight (right) line in segments.
11. Angle between two lines.
12. The parallelism conditions of two lines.
13. . The perpendicularity conditions of two lines.
14. The distance from point to line.
15. Which of the following statements are true and which are false for all lines and
planes in xyz-space: (a) Two lines parallel to a third line are parallel. (b) Two lines
perpendicular to a third line are parallel. (c) Two planes parallel to a third plane
are parallel. (d) Two planes perpendicular to a third plane are parallel. (e) Two
lines parallel to a plane are parallel. (f) Two lines perpendicular to a plane are
parallel. (g) Two planes parallel to a line are parallel. (h) Two planes
perpendicular to a line are parallel. (i) Two lines are either parallel or intersect. (j)
Two planes either are parallel or intersect. (k) A line and a plane either are parallel
or intersect. (l) Two non-parallel lines either intersect or there are parallel planes
that contain them.
Classroom assignments:
1. Define the distance between points А(3;8) and В(-5;14).
2. On axe Ох to find a point with the distance 13 unit from point М(2;5).
3. Show that points А(6;3), В(1;-2), С(-2;-5) belong to one line.
4. The segment bounded by points А(3; -2) and В(6;4), was divided on three
equals parts.
5. Define coordinates of verteces triangle, if it is known the middle its sides
Р(2;3), Q(5;4), R(6;-3).
6. Find polar coordinates of points M 1; 3 , A2 3;2 , B 4;4 , C 2 ; 2 if
the pole coincide with point of origin and polar axe -- with position
direction of absciss(a) axe.
5
7. Find Rectangular coordinates of points: A10; ; B 2; ; C 0; ; D1; .
2
4
10
4
8. Write equation of line passing through the points: а) А(0;2), В(-3;7); б)
А(2;1), В(4;1).
9. Find angular coefficient to line and ordinate of point its intersection with axe
Оу, if it is known line pass through points А(1;1) and В(-2;3).
10. The coordinates of points М1(-3;5) and М2(4;6) are given and equation of
line d 5x-7y=8=0.
To demand:
1. construct line d and points М1 and М2;
2. calculate the distance from point М1 to line d
3. write equation of line passing through point М1, parallel to line d
4. write equation of line passing through point М1, perpendicular to line d
5. write equation of line М1М2;
6. clarify mutual disposition of lines М1М2 and d;if they are not parallel, to
define tangent of angle between them and to find coordinate of points its
intersection.
11. Give equations of (a) the line through (5, 6, 7) and parallel to the line x = 4,
y = 6 – t, z = 9 + 2t, (b) the line through the origin and perpendicular to the plane
3x – 4y + 5z – 18 = 0
12. What are the numbers a and b if a, b, 5 is perpendicular to the plane 4x –2y
+ z = 5?
13. Find the component form of the vector:
a. The vector PQ where
P (1,3) and Q(2,-1) .
b. The vector OP where O is the origin and P is the midpoint of
segment RS , where R (2,-1) and S (4,3) .
c. The vector from the point A (2,3) to the origin.
d. The sum of AB and CD , where A (1,-1) , B (2,0) , C (-1,3) and
D (2,2) .
14. Find a) the direction of P1P2 and b) the midpoint of line segment P1P2 .
a. P1 (1,1,5) and P2 (2,5,0)
b. P1 (0,0,0) and P2 (2,2,2)
Homework: Function. The limit function. Fundamental theorems on limits.
Infinitely small and infinitely large quantities.
Theoretical material: Curves of the second order: circle, ellipse, hyperbola,
parabola.
Solve problems:
1. Vertex of triangle ABC are given: Ax1 ; y1 , Bx2 ; y2 , C x3 ; y3
to define
coordinates of a point of intersection of medians of a triangle.
2. Two vertices of a triangle А(3;8) and В(10;2) and point of intersection of
medians of M (1; 1) are given. Find coordinates of thirds of vertex of a triangle.
3. Make the equation of the line passing through the point of А(-2;1):
а) parallel to the axis Оу;
b) forms an angle 3 with the Ох -axis;
4
c) parallel to the bisector of the first quadrantal angle;
d) perpendicular to the straight line 6x-y +2 = 0;
d) cut off on the y-axis segment length of 5.
4. Find the areas of the triangle with vertices P = (2, 1, 0), Q = (3, 4, 5), R = (6, 1,
2).
5. Find the distance from (2, 4 –1) to the plane z = 2x + y + 3. Give parametric
equations of the line that passes through the origin and through the intersection of
the lines L1: x = 3 + 2t, y = –4t, z = –3 + t and L2: x = 3 + 10t, y = –25 + 5t, z = 4 –
2t.
6. Find the measures of the angles between the diagonals of the rectangle whose
vertices are A (1,0) , B(0,3) , C(3,4) and D(4,1)
7. Let u (3,2) and v (2,5) . Find the a) component form and b) magnitude
(length) of the vector.
3
5
4
5
b). u v .
a). 2u 5v ;
PRACTICAL CLASS № 5.
FUNCTION. FUNCTION LIMIT. FUNDAMENTAL THEOREMS ON
LIMITS. INFINITELY SMALL AND INFINITELY LARGE QUANTITIES.
THE ENDS
Theoretical questions:
1. Definition limit of function in point. 2. Limit of function in infinite. 3. Infinitely
small and infinitely large functions (definition). 4. Basic theorems of limits. 5.
Remarkable limits. 6. Definition and properties of infinitely function. 7.
Comparing the orders of infinitely small. 8. Application of the infinitesimal to the
calculation of limits. 9. One-sided limits. 10. Continuity of function. 11.
Classification of points of discontinuity.
Classroom assignments:
1. Using the properties of limits of functions, find the following limits:
x2 4
; 2)
2
x2 x 5 x 6
2 x 1
4) lim
; 5)
5 x 2
x 1
1) lim
1 x x2
; 3)
lim
2
x 2 x 3 x
x5 2x
;
lim
3
2
x 2 x x 1
lim x
x 1
3
5x 1
;
5x 3
sin 2 3 x
.
lim
2
x 0 sin 2 x
2. Find limits: 1)
sin 5x
; 2)
lim
x 0 sin 3x
3. Find limits: 1)
k
1 , k R ; 2)
lim
x
x
x
lim
x
1 5x ;
x 0
x 3
3) lim
;
x x 2
x
4. Using equivalents infinite smalls, find the following limits:
7
arctg x
sin x 3
1 cos mx
4
1. lim 2
2. lim
3.
lim
2
2 x
e
1
x
4
x
3
x
x
0
x 3
x 0
sin x
2
2 1
ln 1 x
ln cos x
4. lim
5. lim
6. lim
2
2
x
tg 8 x
x
x 0
x 0
x0
5. Investigate continuity of the function and establish character of discontinuity
points:
1 при x 0,
1. f ( x)
0 при x 0
2.
x при x ,
f ( x) sin x при x ,
2
1 при x 2
2 при x 2,
3. f ( x) 4 x 2 при 2 x 2, .
x 2 при x 2
Homework:
Theoretical material: Derivative of function. Differential of function.
Solve problems:
1. Find limits of functions:
3x 2 x 12
а) lim
x
3x 2 x
2
x 2 x 12
3 x 3 x
б) lim 2
в) lim
5x
x 3 x
x 0
5x 6
sin 5 x
г) lim
д) lim 3x 2ln x 3 ln x 4
x
x 0 arctg 3 x
2. Investigate continuity of the function, find discontinuity points and establish
character of discontinuity points.
а)
f x
x4
x 2 3x 4
1 x, x 0
б) f ( x) x 2 1, 0 x 1 .
1, x 1
PRACTICAL CLASS № 6.
THE DERIVATIVE OF THE FUNCTION. GEOMETRIC AND
MECHANICAL MEANING. TABLE OF DERIVATIVES. THE
DIFFERENTIAL OF A FUNCTION
Theoretical questions:
1. Definition of derivative. 2. The geometric meaning of derivative. 3. The basic
rules of finding derivatives. 4. Derivatives of the basic elementary functions.5.
Derivative of composite functions, implicit functions, the functions specified
parametrically, the method of logarithmic differentiation. 6. Application of the
differential to approximate calculations.
Classroom assignments:
1. Find derivatives of following functions:
1) y x 2 e x ;
2) y x 3 arctgx ; 3) y x x 3 ln x 2 ;
arcsin x
sin x cos x
4
; 5) y
; 6) y 2 x 3 5 ;
x
sin x cos x
6
2
7) y tg x ; 8) y cos x ; 9) y sin 2x 3 ; 10) y tg ln x ;
x
ln x
11) y ln x 2 5 ; 12) y ln tg ; 13) y arctg
.
2
3
4) y
2. Calculate derivatives of following functions defined implicitly:
1. x 3 y 2 5xy 4 0
2. arctgy y x 0
3. x 3 y 3x 3 y 2 5 y 3 3x 4 0
4. x 2 y arctg
y
0
x
3. Calculate derivatives of following functions defined parametrically:
x t 3 t
1.
y t 2 t 1
x e t sin t
2.
y e t cos t
x t sin t
x ln t
4.
y 1 cos t
y sin 2t
3.
4. Using the method of logarithmic differentiation calculate derivatives of
following functions:
1. y ctgx
2. y x sin x
3. y cos x arctgx
arccos x
4. y 1 x 2
5. y x arcctgx 6. y tgxcos x
5. Find differential of functions:
1. y arctg x
2. y arcsin x 5
3. y 1 sin 2 x
x2
4. y
arctgx
1 x2
tg
5. y 2
2
x
6. Find the approximate value:
1. 3 26,19
2. 4 16,64 3. 8,76
4. ln 0,9
5. sin 29 0
6. 1,025
7. arctg1,05
Homework:
Theoretical material: L'Hopital's rule. Investigation of the function. Extremum of
the function. Convexity, concavity and inflection points. Asymptote.
Solve problems:
1. Find derivatives of the following functions applying the formulas and rules of
differentiation:
13
2
3 ; б) y 2ctgx 3 sin x ;
4
x
x
x
в) y 19 8 arcsin x ; г) y 3 sin 2 x lg x 3 cos 2 x ;
а) y 5 x
д) y
e x ln x
;
e x ln x
е) y
x2 x 2
.
x3 4
2. Calculate derivatives of functions:
2. y x arctgy
1. 2 y ln y x
4. y x 2 1
3.
Show
x
x at sin t
y a1 cos t
3.
1
5. y x ln x
that
the
1
2
relation y y cos x sin 2 x .
function
y ae sin x sin x 1
satisfies
the
4. Find differentials of functions:
а) y x 1 x 2 ; б) y e x cos x в) y 1 tgx8
5. Using a differential calculate approximately: а) 3 26 ; б) tg 44 0
6. For function y f x а) y xex , y ? б) y ln x, y n ?
PRACTICAL CLASS № 7-8.
ROLLE'S THEOREM, LAGRANGE, CAUCHY. L'HOPITAL'S RULE.
INVESTIGATION OF THE FUNCTION. EXTREMUM OF THE
FUNCTION. NECESSARY AND SUFFICIENT CONDITIONS FOR THE
EXISTENCE OF AN EXTREMUM. CONVEXITY, CONCAVITY AND
INFLECTION POINTS. ASYMPTOTE. THE OVERALL STUDY OF
DESIGN FEATURES
Theoretical questions:
1. L’Hospital’s Rule;
2. Monotonic conditions. Extremum of function;
3. Convexity and concavity. Point of inflection;
4. Аsymptotes;
5. Scheme of investigation of function and charting.
Classroom assignments:
1. Find the limits using L'Hospital's rule:
1. lim
x 0
ln sin 3x
ln x
1
1
lim ln x x 1
2. lim
x2
6.
x 1
x 3 x 10
x 3 3x 2
lim x
x
2
ex
ex 1
4.
lim
x 0 sin x
1x
7. lim x e 1 8. lim x x 9.
x
x 0
3.
lim
x
ln x
x
lim cos x
5.
1
x
x 0
2. Conduct a complete investigation of the functions and construct their graphs: 1.
y
x3
4 x2
2. y
ln
x
3.
Describe the concavity of the graph of f ( x) 2 sin 2 ( x) x 2 for x 0, .
4.Find and so that the function f ( x) x 3 x 2 1 has a point of
inflection at (1;2) .
x 2 3x 1
5.Find the domain and all asymptotes of the following function: y
.
4x2 9
6. Find the maximum and minimum values of f ( x)
Homework:
Theoretical material:
Antiderivative and indefinite integral.
2x 5
on the interval [0, 5].
3
Solve problems:
1. Find the limits using L'Hospital's rule:
1. lim
x 0
ln x
2.
ctg 2 x
1
1
3.
lim
sin x
x 0 x
tgx
lim x
4.
x 0
ex
lim
3
x x
2. Conduct a complete investigation of the functions and construct their graphs: 1.
y
x2
2. y x 2 e x .
2
1 x
x3
.
x2 9
x2 x 2
4. Find the domain and all asymptotes of the following function: y
.
x2
5. Find the maximum and minimum values of f ( x) x2 2 x 4 on the interval [-2,
1].
3. Find the domain and all asymptotes of the following function: y
6. Find the maximum and minimum values of h( x) sec x on the interval , .
6
3
PRACTICAL CLASS № 6.
ANTIDERIVATIVE. INDEFINITE INTEGRAL AND ITS PROPERTIES.
TABLE OF INTEGRALS. DIRECT INTEGRATION, INTEGRATION
WITH THE CHANGE OF VARIABLES AND BY PARTS
Theoretical questions:
1.
Concept of antiderivative and indefinite integral. 2. Main properties of
indefinite integral. 3. Main methods of integration
Classroom assignments:
1. Calculate integrals using table of integrals:
1.
2
x 3 x
dx
x2
2. 2 x 3 x 1 dx
3
3.
1 x 2 dx
x1 x
2
4. cos 2 3xdx 5. tg 2 7 xdx
2. Find integrals using appropriate substitution:
1. x 2 3 3x 2 5dx 2. Sin 3 x Cosxdx 3.
5.
ln 5 x
x dx
4.
x2
dx
x3 1
3x 2
dx
4
3. Find integrals using method of integration by parts:
1. x 2 5 x 7 ln xdx . 2. xe4 x dx . 3. xarctgxdx . 4. ln xdx
5. arcsin xdx .
4. Calculate integrals:
1.
(2 x 3) 3
x 2 dx
4.
e
sin x
dx 5.
cos x
2. cos(7 x 1)dx 3. (5 2 x) 5 dx
e
x4
x 3 dx 6.
a
8 x 4
dx
2x 2
sin 2 x
8.
dx
dx
2
2x 1
cos x
Homework:
Theoretical material: Integration of simple rational fractions. Integration of
irrational functions.
Solve problems:
3
x x5 2
Calculate integrals: 1.
dx 2. e x x 2dx
2
x
3
2
2
( x 3)
dx
3x 1
ln x
dx
dx
3. 4 dx 4.
5.
6. 2
4
1 5x
x x 1
x
x
x2 2 x
tgx 4
7.
dx 8. x 3 5 x ln xdx 9.
dx
sin 2 x
x
7.
x
3
10. x 2 arccosxdx 11. 3x cos xdx
12. x arcsin 2 xdx
PRACTICAL CLASS № 9-11.
INTEGRATION OF SIMPLE RATIONAL FRACTIONS. INTEGRATION
OF RATIONAL FRACTIONS. INTEGRATION OF EXPRESSIONS
CONTAINING TRIGONOMETRIC FUNCTIONS. INTEGRATION OF
IRRATIONAL FUNCTIONS
Theoretical questions:
1. Integration of rational fractions. 2. Integration of expressions containing
trigonometric functions. 3. Integration of irrational functions.
Classroom assignments:
1. Calculate integrals:
dx
dx
( x 3)dx
1.
2. 2
3. 2
x 1
x 6x 7
( x 3)( x 4)
dx
dx
4.
5. 3
9 sin x
sin x
2. Find integrals using appropriate substitution:
( x 2 4 x 4)dx
(4 x 1)dx
dx
2.
3.
x( x 1) 2
( x 1) 2 ( x 2)
x(x 4) 2
dx
5.
( x 1)( x 2) 2 ( x 3) 3
3. Calculate integrals:
1.
3x 1
dx
x 1
1.
x
4.
sin 3 x
cos x 3 dx 5.
2.
2
7. sin 3x cos10 xdx
x
2
xdx
7 x 13
3.
2x
2
4.
x
3
xdx
3x 2
x5
dx
2x 3
cos 3 x
sin 4 x dx 6. cos 4 x cos 7 xdx
sin x cos x
dx
8.
3 sin 2 x
Homework:
Theoretical material: The definite integral. Problems leading to the definite
integral. The Newton-Leibniz formula.
Solve problems:
dx
dx
Calculate integrals: 1.
2.
9 4 sin x
3 cos x
dx
3. sin 5x cos4 xdx 4. sin 4 xdx
5. (1 4 sin x) 3 dx 6. 3
x 1
2
(3x 2 x 1)dx
dx
dx
7. 2 2
8. 2
9.
( x 1) 2 ( x 2)
x x5
x ( x 1)
x 4 dx
x 4 5x 2 4
10.
PRACTICAL CLASS № 12-15
THE DEFINITE INTEGRAL. PROBLEMS LEADING TO THE DEFINITE
INTEGRAL. THE NEWTON-LEIBNIZ FORMULA. APPLICATIONS TO
THE COMPUTATION OF THE INTEGRALS OF PLANE FIGURES
AREAS. CALCULATION THE ARC LENGTH, THE AMOUNT OF BODY
ROTATION. THE IMPROPER INTEGRAL
Theoretical questions:
1. Newton – Leibniz formula. 2. Basic methods of integration.
applications of the definite integral.
Classroom assignments:
1. Find integrals using Newton – Leibniz formula
lg 2
1.
2
0
x
5 dx
x
5
2.
2
dx
2x 3
2
3.
1
x2
dx
3 x
e
4.
1
x x
x x
dx
3. Geometric
1
5.
x4
x 2
0
dx
2. Calculate integrals: 1. 1 x dx x cos t
1 x
0
1
4
2
2. 2 x 2 dx
tg
3.
3
xdx
4.
0
1
x
2
xe dx 5. ln x 4dx
0
2
1
0
3. Find the area bounded by the curve y sin x , the x-axis, and the lines x 0 and
x 2 .
x2 y2
4. Find the area of the ellipse 2 2 1 .
a
b
5. Find the area bounded by the curve x a cos3 t , y b sin 3 t for 0 t 2 .
6. Find the area of the lemniscate r 2 a 2 cos 2 .
7. Find the length of the cissoid r 2a tan sin from 0 to
4
.
Homework:
Theoretical material: Functions of several variables and main properties. Partial
derivatives and differentials.
Solve problems:
1. Find the area bounded by the parabola y 2 2 x 4 and the straight line
x y 2.
2 Find the area bounded by the ellipse x 2 4 y 2 6 x 8 y 9 0 .
3. Find the area of the ellipse whose parametric equations are x a cos t and
y b sin t .
4. Find the parabola y 2 4 x from (0,0) to (-4,4).
PROBLEMS FOR TSIS.
TSIS 1.
ELEMENTS OF LINEAR ALGEBRA
Problem 1. Calculate the determinant
1.
2
1
0
5
3
7
8
9
1
3
5
0
3 0 2 2
4
3
3
5
4 1
11.
1 1
1
2
2 3
1
0
2
3
21.
2 3
2
5
1 1
1
2
2 1 3 2
1 1 3 4
4.
0
1
3
1
0
0
1 2 1
0 1
0
3
0
0
1 5
2 2
2
2
7 8
29
3
6 7
28
2
12.
3 1 1 4
4 1
3
13.
1 2
22.
2
23.
3
0
2
2
11 15
1
1
5
2
2 3
3
2
1 3
3
4
5 1 3
4
4 0 2
3
5
0 0 1 12
2
6 1 1
12
2
1 1
3
8
5
3 5
3
5
1
2
1 2 3
4
3
1
5
12
7
8
3
14.
0
18 28 13
7
5
3
0
3
2 12
2
5
3
8
7
4
3
1 5
0
6
3
8
3
5
4
4
1 4
2
1
24
1
5
16.
5
1
7
7 28 3
4
2
2
6
7
2
3
8
1
5
3
33
10
1
18
15.
9
2 3
8
1 12
6.
5
0
2 3 4 5
5.
1
1
2.
3.
2 1
2
25.
4
3
3 7
1
4
3
1
2
0
1 1
1
5
2
5
0 1
1
2
3
1
1 1
2
3
0
2
1 0 0 2
2 0 2 1
0 2 1 2
2 2 1 3
2 3 1 5
2
0
2 1 2 0
0
1
2
0
26.
1 2 2 1
3 2 2 3
2 1 1 2
7.
3
5
7 2
1
2
3 4
17.
2 3 3 2
1
3
2
1 2 3 0
5 4
27.
0 1 2 3
1 7
0
3 0 1 2
1 4
3
3
3
5 3
8
5 9
0
5
8
2 3 0 1
1 2
8.
2
3
4
1
1 14 3 0
3
3 4
1 2
4
2
3
1 2 1
9.
4
4 0
1 2
2
0
18.
0 20 2 3
3
19.
1 0 2
3 2
1
20
3 5 7
5
2
3
1 4
29.
3
5
1
1
1
4 3 5 0
1 3 5 7
3
3
0
0
2
2 12 0 1
0 1 3 4
3
1
1 3 4 5
3 14 1 2
1
5 2 1
0
1
2
1
10. 4 0 2
1 2 0
5
2 1
28
3
2
0
3
0
4 3 7
1
2
1
1
0
5 5 1
1
2 1 3 5
30.
0
8
4
9
1
5 7 1 3
2
6
7
3
0
1 3 5
1
0
6
5
7
Problem 2. Solve the system of linear equations: 1. by matrix method; 2. by
Cramer's rule:
1.
4 x3 2
3x1
x1 4 x2 3x3 3
2 x x 3x 0
1
2
3
16.
2 x1 x 2 x3 3
x1 3x 2 2 x3 1
x x 5
2
1
2.
3.
4.
4 x1 2 x 2 2 x3 2
3
2 x1 3x 2
x x x 1
2
3
1
4 x1 x 2 2 x3 1
3 x1 2 x 2 3x3 3
x
4 x3 0
1
x1 3x 2 3x 3 2
0
3x1 3x 2
4 x x 2 x 2
1
2
3
3x1 2 x 2 2 x 3 4
4
5. 2 x1 x 2
3 x
3x 3 1
1
1
2 x1 4 x 3
6. 2 x1 2 x 2
4
3x x 3x 3
1
2
3
2 x1 3 x 2 2 x 3 0
7. x1 3x 2 3x 3 4
x 2x 2x 0
1
2
3
3x 3 3
2 x1
8. 4 x1 2 x 2 4 x3 1
x x x 0
1
2
3
9.
17.
x 2 y 3z 8
4 x 5 y 6 z 19
7 x 8 y
1
18.
3x 2 y z 1
6 x 5 y 4 z 2
9 x 8 y 7 z 3
19.
x 2 y 3z 14
2 x y z 1
3x 2 y 2 z 13
20.
2 x 3 y z 8
5 x y z 10
x 3y 4z 3
2 x 3 y z 0
21. 5 x y 2 z 1
x y z 3
22.
2 x 3 y z 6
3x y z 1
5 x 2 y 4 z 11
23.
3x y 2 z 13
2 x y z 0
5 x 3 y 7 z 28
3x1 2 x2 4 x3 2
3x2 3x3 4
x
3x3 2
1
24.
x1 x 2 4 x 3 2
4 x 3 1
x1
2 x 2 x 3x 2
1
2
3
25.
10.
2 x1 2 x 2 2 x3 3
3
3x
x3 2
1
26.
x1 x 2 4 x 3 2
x 2 4 x 3 1
2 x 2 x 3x 2
1
2
3
27.
11. 4 x1 x 2
12.
x1 2 x2 3x3 6
13. 4 x1 5x2 6 x3 9
7 x 8 x 6
2
1
14.
x1 2 x2 3x3 6
4 x1 5 x2 6 x3 15
7 x 8 x 9 x 24
2
3
1
3x y 2 z 10
z 22
7 x
x 3 y 2 z 2
x 3y z 3
2 x y 4 z 5
3x 2 y 5 z 10
2 x y 4 z 7
7 x 3 y z 3
5 x 2 y 3z 4
2 x 3 y 4 z 1
3x y 2 z 12
4 x 3 y 3z 9
5 x1 8 x2 x3 2
28. 3x1 2 x2 6 x3 7
2 x x x 5
2
3
1
2 x1 3x 2 x3 7
29. x1 4 x2 2 x3 1
x 4x
5
2
1
15.
x1 2 x2 3x3 3
2 x1 6 x2 4 x3 6
3x 10 x 8 x 21
2
3
1
5 x 8 y z 2
30. 3x 2 y 6 z 7
2 x y z 5
Problem 3. Investigate systems on compatibility and, in case of compatibility,
solve them by Gauss's method:
3x1 x2 8 x3 2 x4 1
1. 2 x1 2 x2 3x3 7 x4 2
x 11x 12 x 34 x 5
2
3
4
1
7 x1 2 x 2 x3 2 x 4 2
2. x1 2 x 2 x3 x 4 1
2 x 5x 2 x x 1
2
3
4
1
3.
x1 x 2 10 x3 x 4 1
5 x1 x 2 8 x3 2 x 4 2
3 x 3x 12 x 4 x 4
2
3
4
1
4.
12 x1 x 2 7 x3 11x 4 1
x1 10 x 2 3x3 2 x 4 1
4 x 19 x 4 x 5 x 1
2
3
4
1
x1 2 x 2 x 3 4 x 4 1
5. 2 x1 x 2 3x 3 x 4 5
x 3 x x 6 x 1
2
3
4
1
2 x1 x 2 3x3 x 4 1
6. x1 5 x 2 x3 x 4 2
x 16 x 6 x 4 x 7
2
3
4
1
7.
8.
6 x1 9 x 2 21x3 3x 4 12
4 x1 6 x 2 14 x 3 2 x 4 8
2 x 3x 7 x x 4
1
2
3
4
2 x1 x 2 2 x3 x 4 1
x1 10 x 2 3x3 2 x 4 1
4 x 19 x 4 x 5 x 1
2
3
4
1
5 x1 2 x 2 3x 3 4 x 4 1
9. x1 4 x 2 3x3 2 x 4 5
6 x 2 x
2 x 4 6
2
1
16.
17.
18.
19.
20.
21.
22.
23.
24.
x1 2 x2 2 x3 3x4 0
6 x1 3x2 3x3 x4 0
7 x 9 x 9 x 10 x 0
1
2
3
4
2 x1 x 2 x3 x 4 1
2 x x
3x4 2
1
2
x 3 x 4 3
3 x1
2 x1 2 x 2 2 x3 5 x 4 6
2 x1 x 2 x3 x 4 1
3 x 2 x 2 x 3 x 2
1
2
3
4
5 x1 x 2 x3 2 x 4 1
2 x1 x 2 x3 3 x 4 4
x1 2 x 2 3 x3 4 x 4 4
x 2 x 3 x 4 3
3x4 1
x1 3 x 2
7 x
3 x 3 x 4 3
x1 2 x 2 3 x3 4 x 4 11
2 x 3 x 4 x x 12
1
2
3
4
3
x
4
x
x
2
x
2
3
4 13
1
4 x1 x 2 2 x3 3 x 4 14
x1 2 x 2 3 x3 3 x 4 2
2 x x x x 1
1
2
3
4
3 x1 x 2 2 x3 2 x 4 3
x1 3 x 2 4 x3 4 x 4 1
2 x1 x 2 x3 x 4 6
x 2x 2x x 1
1
2
3
4
3 x1 2 x 2 x3 2 x 4 2
x1 3 x 2 3 x3 2 x 4 1
x1 x 2 x3 x 4 2
2 x x 3 x x 1
1
2
3
4
3
x
2
x
2
x
3
2
3
1
x1
4 x 3 2 x 4 3
x1 3 x 2 3 x3 x 4 0
2 x x x x 1
1
2
3
4
3
x
2
x
x
2
x
2
3
4 3
1
x1 x 2 2 x3 2 x 4 1
10.
3x1 x2 8 x3 2 x4 1
2 x1 2 x2 3x3 7 x4 2
x 11x 12 x 34 x 5
2
3
4
1
x1 3x 2 x3 12 x 4 1
11. 2 x1 2 x 2 x3 10 x 4 14
3 x x
x4 0
2
1
12.
3x1 x 2 8 x3 2 x 4 1
x1 11x 2 12 x3 34 x 4 5
x 5 x 2 x 16 x 3
2
3
4
1
13.
x1 x 2 x 3 4
x1 2 x 2 3 x 3 0
2 x
3 x 3 16
1
14.
x1 2 x2 2 x3 3 x4 1
6 x 3 x 3 x x 9
1
2
3
4
7
x
x
x
2
x
1
2
3
4 8
3 x1 9 x2 9 x3 10 x4 12
x1 x2 x3 4
15. x1 2 x2 3x3 0
2 x
2 x3 3
1
25.
2 x1 x 2 x3 x 4 1
2 x x
3x4 2
1
2
x 3 x 4 3
3 x1
2 x1 2 x 2 2 x3 5 x 4 6
26.
7 x1 5 x 2 2 x3 4 x 4 8
3 x 2 x x 2 x 3
1
2
3
4
2
x
x
x
2
x
1
2
3
4 1
x1
x3 24 x 4 1
27.
3 x1 2 x 2 3 x3 5 x 4 10
2 x x 5 x x 5
1
2
3
4
x1 x 2 3 x3 2 x 4 2
2 x1 2 x 2 x3 x 4 1
x1 x 2 6 x3 4 x 4 6
3 x x 6 x 4 x 2
2
3
4
28. 1
2
x
3
x
9
x
2
x
2
3
4 6
1
3 x1 2 x 2 3 x3 8 x 4 7
29.
3 x1 7 x 2 7 x3 2 x 4 22
x 8 x 10 x 3 x 35
1
2
3
4
4
x
7
x
14
x
5
x
2
3
4 48
1
x1 2 x 2 3 x3 x 4 12
30.
3 x1 7 x 2 7 x3 2 x 4 8
x 8 x 10 x 3 x 3
1
2
3
4
4
x
2
x
3
x
x
17
2
3
4
1
5 x1 17 x 2 x3 2 x 4 24
TSIS 2.
ANALYTIC GEOMETRY IN THE PLANE.
Problem 1. Coordinates of points М1, М2 and the equation of straight line d are
given. It is required:
1. construct a straight line d and points М1 and М2;
2. Calculate the distance from the point М1 to the straight line d;
3. Write the equation of the straight line passing through the point М1, parallel to
the straight line d;
4. write the equation of the straight line passing through the point М2 is
perpendicular to straight line d;
5. write the equation of the straight line М1 М2;
6. determine a relative position of straight lines М1 М2 and d; if they are not
parallel, to determine the tangent of the angle between them and find the
coordinates of the point of their intersection.
1. d : x y 1 0, M1 2;3 M 2 3;1
2. d : y 2 0, M1 1;1 M 2 0;3
3. d : x y 1 0, M 1 1;0 M 2 2;1
4. d : 2 x 2 y 1 0, M1 2;3 M 2 2;2
5. d : 2x y 0, M 1 1;2 M 2 3;2
6. d : x y 1 0, M 1 2;3 M 2 1;1
7. d : 2 x 2 y 1 0, M 1 2;3 M 2 3;1
8. d : x y 4 0, M 1 0;3 M 2 4;1
9. d : x y 1 0, M 1 0;3 M 2 2;1
10 d : x 3 0, M1 2;3 M 2 3;1 .
11. d : x 2 y 6 0, M 1 2;2 M 2 1;1
12. d : x y 1 0, M1 1;1 M 2 3;1
13. d : x 3 y 3 0, M 1 2;0 M 2 0;1
14. d : y 1 0, M 1 2;3 M 2 4;1
15. d : 4 x 1 0, M 1 2;2 M 2 1;1
16. d : x 4 y 6 0, M 1 0;3 M 2 3;1
17. d : x 5 y 1 0, M 1 2;3 M 2 0;1
18. d : 3x 2 y 6 0, M 1 2;4 M 2 3;1
19. d : 3x 2 y 6 0, M 1 2;3 M 2 3;2
20. d : 2x 6 y 12 0, M 1 2;3 M 2 3;1
21. d : 3x 5 y 15 0, M 1 2;2 M 2 3;1
22. d : 2x 3 y 12 0, M 1 1;3 M 2 3;1
23. d : 3x y 4 0, M1 2;0 M 2 0;1
24. d : x 5 y 1 0, M 1 2;3 M 2 3;1
25. d : x 4 y 6 0, M1 0;4 M 2 3;1
26. d : 3x 7 y 21 0, M1 2;1 M 2 3;1
27. d : x 5 y 10 0, M 1 2;3 M 2 4;1
28. d : 3x y 0, M 1 2;1 M 2 3;5
29. d : x 4 y 0, M 1 1;3 M 2 1;4
30. d : 3x 7 y 21 0, M1 2;3 M 2 3;6
Problem 2. The equation of a curve of the second order is given. Find lengths of
half-axles, coordinates of focuses, eccentricity, the equations of asymptotes (for
hyperbole). Construct this curve.
1. x 2 y 2 16
2. 4x 2 y 2 16
3. 9x 2 y 2 9
4. x 2 64 y 2 16
5. x 2 9 y 2 36
6. 4x 2 16 y 2 64
7. 4x 2 25y 2 100
11. 5x 2 20 y 2 80
12. 16x 2 y 2 64
13. 9x 2 4 y 2 36
14. x 2 y 2 1
15. x 2 4 y 2 4
16. 25x 2 4 y 2 100
17. 4x 2 y 2 1
21.
22.
23.
24.
25.
26.
27.
x 2 25y 2 100
4 x 2 y 2 16
x2 y2 1
x2 y2 9
x 2 4 y 2 16
9x 2 y 2 9
4x 2 y 2 1
8. x 2 4 y 2 36
18. 4x 2 y 2 1
28. x 2 y 2 4
9. x 2 4 y 2 64
19. 4x 2 9 y 2 36
29. x 2 9 y 2 9
10. 4x 2 9 y 2 36
20. x 2 4 y 2 1
30. x 2 4 y 2 1
Problem 3. Constitute a canonical equation of the parabola whose vertex is at the
origin, if it passes through point M and symmetric given axis. Find the coordinates
of the focus and the equation of the directrix. Construct this parabola.
1. М(2,2), axis ОХ
16. М(2,2), axis ОУ
2. М(5,-3), axis ОХ
17. М(1,-3), axis ОУ
3. М(-1,-2), axis ОХ
18. М(-2,-2), axis ОУ
4. М(3,-3), axis ОХ
19. М(2,-3), axis ОУ
5. М(2,3), axis ОХ
20. М(-4,3), axis ОУ
6. М(-2,-1), axis ОХ
21. М(-1,-1), axis ОУ
7. М(1,1), axis ОХ
22. М(-2,1), axis ОУ
8. М(-2,-2), axis ОХ
23. М(-1,-2), axis ОУ
9. М(-4,6), axis ОХ
24. М(-2,3), axis ОУ
10. М(3,-5), axis ОХ
25. М(-3,-1), axis ОУ
11. М(1,4), axis ОХ
26. М(-1,-2), axis ОУ
12. М(-2,3), axis ОХ
27. М(3,2), axis ОУ
13. М(-4,-2), axis ОХ
28. М(-3,-4), axis ОУ
14. М(-4,5), axis ОХ
29. М(4,-3), axis ОУ
15. М(1,-3), axis ОХ
30. М(1,2), axis ОУ
TSIS 3.
FUNCTIONS.LIMITS. CONTINUITY.
Problem 1. Find limits of functions
1. а)
3x 3 5 x 2 2
lim
x 2 x 3 x 2 x
б)
2 x 2 3x 1
lim
2 5x 3
x 1 2 x
2x 3 3
в)
г) lim
lim x 2 1
д) lim 2 x 1ln x 3 ln x
x 0
x 3
arctg 2 x
4x
x
2. а)
в)
3x 2 5 x 2
lim
x 2 x 2 x 3
б)
3x 2 2
2x 5 3
г)
lim
x 2
x 5
2 x 4 5x 2 3
3. а) lim
x 5 x 4 2 x 3 4 x
в)
lim
x 1
2x 3 1
5 x 2
lim
x 2 14 x 5
x 2 2 x 15
x 2 ctg 2 x
lim
sin 3x
x 0
д)
x 3
lim
x x 2
x
x2 x 2
б) lim
2 x 1
x 1 2 x
cos x cos 3 x
г) lim
x 0
x2
2x 1
д) lim
x 2 x 1
x
4. а)
в)
5 x 2 3x 1
lim
x 3x 2 x 5
1 x 4
lim
x 5 2 2 x 6
x 2 7 x 10
lim
2 9 x 10
x 2 2 x
б)
г)
cos 3x cos 5x
lim
x 0
x2
4x3 2x 1
5. а) lim
x 2 x 3 3 x 2 x
2x
2x 2 7x 4
lim
2 13x 20
x 4 2 x
б)
3 x 11
в)
4x 1
x 4 x
д) lim
sin x
г) lim
lim 2 x 6
arcsin x
д) lim x 5ln x 3 ln x
x 0
x 2
x
6. а)
в)
3 7 x 2 5x3
lim
x
2 2 x x3
lim
x 5
б)
x 3
9 x 2
4 x 3
lim
x 9
в)
г)
x 4
д)
lim 3x 5
2
arcsin 2 x
lim
5x
x 0
cos x cos 3 x
г) lim
xtg3x
x 0
8 x 2
x2 7 3
lim 1 2 x
д)
г)
1 1 x2
lim
x 0 cos x cos 2 x
x 2
б)
lim
x 5
1 x 3
x 2 25
x 2 8 x 15
6x
г) lim
lim 2 x
arcsin 3x
д) lim 2 x 7 ln x 4 ln x
x 4
x 0
x
3x 3 x 2 x
11. а) lim
x
3x 2 x
2x 1
lim
x 2 x 1
3x 5
д) lim
x 3 x 1
x 2
в)
x
x 0
x 2 4 x 21
lim
2 7x 3
x 3 2 x
б)
3x 14 x 2
10. а) lim
x 7 x 2 x 2
1
3 x 1
2x 2 7x 4
б) lim
2 13x 20
x 4 2 x
3 2 x 5x 4
lim
x 2 2 x 2 x 4
lim
д)
3x 2 5 x 2
б) lim
2 4x 3
x 1 x
3x 4 2 x 2 7
8. а) lim
x
9x 4 x 5
9. а)
lim 1 cos 2x
x 0
x 3
2x 2 4
2 5 x
в) lim
x 1 3 8 x
1 cos 4 x
г)
6 x 3 3x 2 3
7. а) lim
x
3 2 x3 x
в)
lim
x 2 10 x 21
x 2 8 x 15
б)
x2 x 6
lim
2 x 21
x 3 2 x
2x
в)
3x 2 1 2
x x2
lim
x 1
д)
5x
г)
lim 3arctg 3x
x 0
lim x 2ln 2 x 3 ln 2 x 1
x
4 x5 2 x 1
12. а) lim
x 8 x 5 3 x 2 x
б)
5 22 x
в)
2 x 2 x 10
lim
x 2
x2 x 2
1 cos 2 x
г) lim
lim 1 x 4
2 x tgx
д) lim 3x 1ln 2 x 1 ln 2 x 1
x 0
x 3
x
6x3 2x 7
13. а) lim
x 3 x 3 2 x 2 x
в)
lim
x 3
д)
2 2x 2
2 x 1
lim 2x 5
2x
г)
б)
lim
x2
x2 x 2
x2 x 6
lim x tgx ctg
2
2x
x 0
x 3
x 3
7 x 4 2x3 2
14. а) lim
x
x4 x
3x 2 x 2
б) lim
2 4x 1
x 1 3 x
1 3x 2 x 6
в)
1 cos 4 x
г) lim
lim
x 5x
cos 2 x 1
д) lim x 3ln x 1 ln x 2
2
x 5
x 0
x
15. а)
в)
5x3 2 x 1
lim
x x 3 3 x 2 x
lim
5 6x 1
г)
2 x
б)
x 2 2 x 15
lim
2 7 x 15
x 5 2 x
x sin x
lim
cos x cos 3 x
д) lim x 2ln 3 2 x ln 4 2 x
x 4
x 0
x
4x 2 2x 1
16. а) lim
б)
x 2 x 3 2 x 2 x
в)
2 x 1
2x 3
г)
lim 3
x 3
lim
x 0
5 6 x 4 x3
17. а) lim
x 2 x 3 3 x 2 x
x2
в) lim
x 2 2 2 x
18. а)
x 2 7 x 12
lim
2 13x 20
x 4 2 x
1 cos 2 x
x tg3x
д)
lim 2x 3
x
x 2
2 x 2 5x 7
б) lim
2 x2
x 1 3 x
д) lim 2 x 5 x 9
2x
г)
lim x ctg5 x
x 0
3 x 5x 4
lim
x 2 2 x 2 3x 4
б)
2
x 3
x 2 3x 2
lim
2 5x 2
x 2 2 x
x 2
в)
lim
x 0
9 x2 3
x 2 25 5
1 cos 3x
г)
б)
lim 3 2 x
2x
x 1
x 0
3x 14 x 2
19. а) lim
x 7 x 2 4 x 2
3x 2 5 x 50
lim
2 8 x 15
x 5 x
1 x 3
в)
д)
lim x sin 2 x
sin 6 x
г) lim
lim 2 x 6
sin 3x
д) lim 2 3x ln 3x 1 ln 2 3x
x 0
x 2
x
20. а)
1 4 x3 2 x
lim
x 8 x 3 3x 2 x
б)
2 x 2 x 10
lim
x 2
x 2 2x
sin 2 x
lim
x 0 2 x tgx
x4
г)
lim
x 4 5 5 x 5
д) lim 4 xln 2 x 1 ln 2 x 1
в)
x
21. а)
в)
3x 2 x 12
lim
x
3x 2 x
lim
x 0
д)
x 2 2 x 12
lim
2 5x 6
x 3 x
б)
3 x 3 x
5x
г)
sin 5 x
lim arctg3x
x 0
lim 3x 2ln x 3 ln x 4
x
22. а)
в)
1 x x3
lim
x x 2 3x 3
lim
x 7
23. а)
2 x 3
x7
б)
2x 1
lim
x 2 x 1
arcsin 3x
г) lim
д)
6x
x 0
2x3 x 2 5
lim
x
x3 x 2
б)
x x
в)
x 2 3x 2
lim
2 5x 2
x 2 2 x
3 x 1
x 2 7 x 10
lim
2 9 x 10
x 2 2 x
1 cos 2 x
г) lim
lim x x
x
д) lim 3x 2ln 3x 1 ln 3x 2
2
x 0
x 0
x
24. а)
в)
3x 4 x 3 6
lim
x
2x 4 x 1
lim
x 0
x
1 3x 1
г)
б)
x 2 ctgx
д)
lim
x 0 sin 3x
2x 2 6x 5
25. а) lim
x 1 x 5 x 2
в)
lim
x 3
2x 2 2
x 1 2
3x 2 5 x 2
lim
2 4x 3
x 1 x
lim 5 2x x 4
2x
2
x
x 2 4 x 21
б) lim
2 7x 3
x 3 2 x
г)
lim
x 0
1 cos 2 x
x sin 2 x
x 1
д)
lim x 2ln 3x 2 ln 3x 1
x
5x 4 x 3
26. а) lim
x x 4 12 x 1
б)
3x 2 14 x 5
lim
2 2 x 15
x 5 x
1 3x 1 2 x
г)
lim
x x2
x 0
д) lim 2 x 1ln x 3 ln x
в)
x 2 ctg3x
lim
sin 2 x
x 0
x
27. а)
в)
x 2 x 2 3x 5
б)
lim
2 3x x 5
x
1 3x 2 1
г)
lim
lim
x4
x 2 x 12
x 2 2x 8
1 cos 6 x
lim 1 cos 4x
x 0
x3 x 2
д) lim x 5ln x 3 ln x
x 0
x
28. а)
в)
5 x 2 3x 1
б)
lim
3x x 2
x
lim
x 3
29. а)
x2 x 2
lim
2
x 1 2 x x 1
x
tg 2
x
2x 1 5
2 д)
3 x 3
г) lim
7
6
x
lim
x 3
x 0
x 1
x2
7x 4 2x3 3
б)
lim
3 x2 x4
x
1 3x 2 x 6
в) lim
г)
x 2 5x
x 5
8 x 5 3x 2 9
30. а) lim
б)
2x 2 4x5
x
в)
lim
x 9
x 3
2x 2 4
г)
lim
x 1
x 2 25
x 2 8 x 15
1 cos 4 x
д)
lim
x 0 2 x tg 2 x
3x 2 5 x 2
lim
x 2 4x 3
x 1
4x 1
lim
x 4 x
lim 5 x ctg3x д) lim 3x 8
x 0
2 x 1
2
x 3
x 3
Problem 2. Investigate functions on a continuity, find points of discontinuity (if
they are). Establish character of each point of discontinuity.
1. а) f x
x 1
x 2 3x 2
1 x, x 1
б) f ( x) x 2 1, 1 x 2
2 x 3, x 2
2. а) f x
x2 1
x 1
16. а) f x
x4
2 x 2 3x 20
x , x 0
2
б) f ( x) cos x, 0 x 2
x 2 , x 2
16 x 2
f
x
17. а)
x4
x, x 0
б) f ( x) 1 x, 0 x 1
1
, x 1
1 x
x2
3. а) f x
x 2 3x 2
1
x , x 1
б) f ( x) x, 1 x 2
3, x 2
x2 x 2
4. а) f x
3x 3
2 x, 0 x 1
б) f ( x) 2 x, 1 x 2
0, x 2
5. а) f x
б)
x 2 4x 3
x 1
x , x 0
f ( x) tgx, 0 x
4
2, x 4
6. а) f x
x4
x 2 3x 4
x 2, x 0
б) f ( x) 1 x 2 , 0 x 1
x 1, x 1
7. а) f x
x2
x2 4
x , x 0
б) f ( x) Sinx , 0 x
x 2, x
x2 4
8. а) f x
x2
2 x, x 0
б) f ( x) x , 0 x 4
1, x 4
x, x 0
б) f ( x) 2, 0 x 2
x, x 2
18. а) f x
x3
x 2 3x 18
x , x 0
б) f ( x) Sinx , 0 x
x 2, x
x 2 4 x 21
19. а) f x
x3
2 x , x 0
б) f ( x) x 2 1, 0 x 1
2, x 1
20. а) f x
x 2 4x 4
4 2x
1 x, x 0
б) f ( x) x 2 , 0 x 2
1
x 3, x 2
2
x4
21. а) f x
x 2 3x 4
1 x, x 0
б) f ( x) x 2 1, 0 x 1
1, x 1
22. а)
б)
f x
x
x2 x
(1 x), x 1
f ( x) ( x 1) 2 , 1 x 0
x, x 0
23. а)
f x
x4
x 2 3x 4
2 x, x 1
б) f ( x) x 2 1, 1 x 1
x 3, x 1
9. а)
f x
x2
2
x 3x 10
cos x, x 0
б) f ( x) x 2 1, 0 x 1
x, x 1
10. а) f x
б)
f x
3x 2 12
x2
x , x 0
2
f ( x) x 1 , 0 x 2
x 3, x 2
12. а) f x
б)
x 2 3x 10
Sinx , x 0
f ( x) 2, 0 x 1
1
, 1 x 4
x 1
11. а)
б)
5 x
3 x
x 2 x 12
x 2 , x 0
f ( x) tg 2 x, 0 x
4
2, x 4
13. а) f x
1 x
x 2 3x 4
2 x, x 0
б) f ( x) x , 0 x 4
1, x 4
x2 9
14. а) f x
x3
cos x, x 0
б) f ( x) 2, 0 x 1
1
, 1 x 4
x
x6
15. а) f x
x 2 3x 18
x2 9
x3
4 x, x 1
б) f ( x) x 2 2, 1 x 1
2 x, x 1
24. а)
f x
25. а)
5 x 2 3x
f x
x
б)
x , x 0
f ( x) cos x, 0 x
2
x 2 , x 2
x3 x 2
2x 1
x , x 0
б) f ( x) x 2 , 0 x 2
x 1, x 2
26. а)
f x
1
27. а) f ( x) 2 1 x
2
1
x , x 1
б) f ( x) x, 1 x 2
3, 2 x 3
x2 1
x 1
x
,
x0
б) f ( x) Sin 2 x, 0 x
x 2, x
28. а) f ( x)
29. а) f ( x) e
б)
x
1
x
x, x 0
f x cos x, 0 x
1
x , x
30. а) f x
x 2 3x 2
x 1
x 2 1, x 0
б) f ( x) 1 x, 0 x 2
2, x 2
б)
1
x , x 0
f x 1,
0 x 1
x,
1 x 3
TSIS 4.
THE DERIVATIVE OF A FUNCTION.
APPLICATION OF THE DERIVATIVE.
Problem 1. Find derivatives of functions.
1.1.
б) y e cos x 3
3
а) y 2 4 x 3
2
x x 1
t
y
x cos
2x
в) y x
г) tg 5 x д)
2
x
y t sin t
4 sin x
1.2. а) y x 2 1 x 2
б) y 2
cos x
3
x t 8t
г) x y arctgy 0
д)
y t 5 2t
1.3. а) y x
3
1 x2
1 x
1
tg 2 x
x t sin t
д)
y 1 cos t
б) y
г) y sin x cosx y
3 6x
1.4. а) y
в) y x
1
x
в) y x ln x
2
б) y sin 2 x x cos x в) y x tgx г)
y
x
arctg
x
y
3 4 x 5x
x e 3t
д)
y cos 3t
y
x
sin 2 x
ln x
2
x
1.5. а) y 2 2
б) y
в)
г)
y
arctgx
y
x
e
2
2
3
cos
x
a x
2
x 3 cos t ]
y 2 sin 3 t
1
1.6. а) y 2
б) y 2tg 3 x 2 1
3
5
x 1 5 x 1
x 3 cos t
в) y arctg 2 x x г) e x 1 e y 1 1 0 д)
2
y 4 sin t
2
1 x2
1.7. а) y
1 x2
x 3t t 3
д)
y 3t 2
3
д)
1
2
б) y tg 2 x ln cos x
в) y x x 2 г) x 3 y 3 3axy 0
x
1.8. а) y 33 x 5 5 x 4
б) y arctg tg 2 x
5
x
x t ln cos t
y t ln sin t
в) y sin x ln x г) x y a sin y 0
1.9. а) y 55 x 2 x
1
x
б) y
д)
arcsin x
1 x
в) y cos x x
2
2
г) ln y arctg
x
y
x ln t
д) 1 1
y 2 t t
1.10. а) y x 2 1 3 x 3 1 б) y arctg
в) y sin 2 x
y 2t t 2
д)
y 3t t 3
г) x y e arctgx 0
x
3 x
x2
y
2
1.11. а) y x x
б) y 5
3
в) y x
arctg2 x
ln 2 x
г) e x y xy 0
x 2 cos 3 t
д)
y 4 sin 3 t
1.12. а) y 3
3
x 2 3x 1
2 6 x 5 б) y ln arcctg 2 x
x cos t t sin t
y sin t t cos t
в) y x x
г) e xy x 3 y 0 д)
1.13. а) y 3
x
1 x
б) y sin 3 5x Cos 5 3x в) y cos x ln x г) y sin x y x 0
3
x 2t 3 t
д)
y ln t
1.14. а) y
3
x
1 x
1 x
б) y ln arcsin 2 x
в) y cos 2 x tgx
г) tgx 2 y 3x y 0
x ctgt
д)
1
y cos 2 t
1.15. а) y x 2 3 x
б) y e tg 2 x cos x
в) y x x 2
3x
x ln t
1 1
y 2 t t
1.16. а) y
в) y arctgxx
x x
x x
2
б) y ln e x 1 e 2 x
г) x y 2 x 3 y 3 0
x arcsin t 2
д)
t
y
1 t2
г) y x y xy 0 д)
1
1 cos x
б) y ln
1 cos x
x
x arctgt
x
г) ln x 2 y 0 д)
1 t2
y
y ln
t 1
1.17. а) y 5 x 2 x
в) y sin 3x cos x
2
1 x
б) y arctge 3 x
1.18. а) y 1
в) y x
1
x
x arcsin t
г) arccosx y x 2 y 0 д)
y 1 t
x3 1
1.19. а) y 3
3x 2
x 2 1
1
в) y 1
x
б) y arccostg2x
x t t 2 1
x
г) cos x 2 y 0 д)
1 1 t2
y
y ln
t
4
arcsin 2 x
1.20. а) y 4 x 2
б) y
x
1 4x 2
x
в) y cos 5x
x
x 1 t 2
г) tgx y xy 0
д)
t
y
1 t2
3
4 sin x
1.21. а) y 2 4 x 3 3
б) y
в) y x 3 ln x
cos 2 x
x x 1
2
x 4 cos t
г) x y e y arctgx 0 д)
y 8 sin 3 t
1
1
1.22. а) y 2
б) y tg 3 x ln cos x
3
3
3
x 1 5 x 1
в) y cos x x г) ln y arctg
1.23. а) y 3 x 2 x
y 9t t 2
y
д)
x
y 6t t 3
б) y ln arctg 2 x
в) y cos x ln 3 x
x 2 ln t
г) tgx 2 y 3x y 0 д) 1 1
y 5 t t
1.24. а) y
x x
x x
б) y ln
1 cos x
1 cos x
в) y x
x 1 t 2
д)
t
y
4 1 t2
1
1.25. а) y x 2 6 1 x 2 б) y 5
в) y x 3tgx
tg 2 x
x
г) cos x 2 y 0
y
x2 2
г) yx 2 e
x cos 2 t
y
д)
x
y 4 sin 2t
1.26. а) y 5
1 x2
1 x2
б) y arctgtg7 x в) y cos 2 x x
x 2 cos 4 t
д)
y 9 sin 2 t
г) x y e arctgx 0
y
5
1.27. а) y 5
x 3x 1
2
1
1.28. а) y 5 x x
x
б) y arctge
2
x
a2 x2
x ln t
д) 1 1
y 2 t t
г) x y a sin y 0
3
x 3x 1
2
г) e x y
2
x
б) y 2tg 3 x 2 1 в) y x x 2
x
в) y arctgxx
1
в) y 1
x
3x
t
x cos
д)
2
y t sin t
г) tgx y xy 0
1.30. а) y 3
б) y sin 5 3x Cos 3 5x
2 6x 5
x arcsin t 3
г) y x y xy 0 д)
t
y
1 t2
в) y cos x tg 2 x
1.29. а) y
3
б) y e tg 2 x cos x
2 6x 5
x arcsin t 2
xy 0 д)
t
y
1 t2
Problem 2. Find limits of functions, using L'Hospital's rule.
e ax e 2 ax
2.1. а) lim
ln 1 x
x 0
2.2. а)
1 2x 1
2 x x
lim
e e 2x
x sin x
x 0
2.4. а)
б)
б)
б)
lim 4 2 xe
tgx
lim
Sin3x
lim
2
x 0 ln Cos 2 x x
б)
4
2 x
lim cos x ln 2 x
x
2
2
x 0
2
ln 1 x 2
lim
x
x 0 cos 3 x e
2.5. а)
cos x
x 0
x
x
lim 2 x
x
lim
3
x 1
2.3. а)
б)
2
lim e
x
x
x
1
x
1
2.6. а)
б)
e x 1
2
lim 2arctgx
x
2
2.9. а)
lim
x 0
2.10. а)
tgx x
x sin x
1 4Sin
lim
2 x
6
1 x
xCosx Sinx
2.11. а) lim
x3
x 0
2.12. а)
б)
ln x
e
lim x
x
2.14. а)
lim
x 1
x 1
1
б) lim
x 0 x
б)
2.17. а)
lim
x 0
2.18. а)
2.19. а)
2.20. а)
2.21. а)
2.22. а)
tgx
lim Cos
x
б)
3
4
2x
tg 2 x
3
xa
ctgx a
a
lim arcsin
lim 2arctgx ln x
x
5
1 x
1 Sin
x
4
ctg
б)
2
sec 2 x 2tgx
lim 1 cos 4 x
x
4
x
tg
lim 3 2 x 2
x a
x
ln x
2.15. а) lim 3
x
x
2.16. а)
ctg 2 x
1
x 0
2.13. а)
lim cos x
x
2
б) lim arccos x
x 0
б)
lim ctgx
1
1
x
x 0
б)
2
x 1
x 0
ln x 2 3
lim
2
x 2 x 3 x 10
a ln x x
2.8. а) lim
ln x
x 1
2.7. а)
1
lim x e
x
2
ln sin mx
lim
ln sin x
x 0
x
1 4 cos 2
3
lim
x
x 1
cos
2
3
3
x a
lim
x a
x a
x
e e 2 x
lim
x 0 ln 1 x
e ax e 2 ax
lim
ln 1 x
x 0
1x
a 1 x
lim
x
a 0
ctg 2 x
x 0
1
x
б) lim
ln x
x 1 ln x
б)
1
lim x
2
1
e x 1
б) lim
2
x 2arctgx
б)
2
x
lim 1 xtg
2
x 1
б)
lim arcsin x ctgx
x 0
б)
x
1
lim x 1 ln x
x 1
б)
2 x
lim
cos x
x
б)
2
lim 4 2 xe
x 0
4
2.23. а)
2.24. а)
Sin3x
lim ln Cos2 x
x 0
2.25. а)
lim
x 1
2.26. а)
a ln x x
ln x
lim
1 x
x
б)
2
1
lim x e
x 0
1
1
x
lim 3 2 x
tg
x
2
x 1
x
6
б)
ex
lim
5
x x
lim
lim Cos
x
2
ln x
3
x
x
2.29. а) lim 4
x
x 0
ctg
2
tgx x
2.30. а) lim
x 0 x sin x
2.28. а)
2
2
lim cos x ln 2 x
x
б)
1 4Sin 2
x 1
2.27. а)
б)
e x e x 2x
lim
x sin x
x 0
3
4
2x
tg 2 x
3
1x
б) lim a 1 x a 0
x
1
x
б) lim
ln x
x 1 ln x
x
б) lim 1 x tg
2
x 1
б)
x
1
lim x 1 ln x
x 1
Problem 3. Conduct a complete investigation of the functions and construct their
graphs
x3
3.1. а) y 2
x 2x 3
б) y x ln( x 2 4)
8x
x2 4
x
2
б) y ln
x2
3.2. а) y
3 x 2 7 x 16
x2 x 6
б) y ln( 1 x 2 )
3.3. а) y
x3 4
x2
б) y 4 x e x3
3.4. а) y
x3 8
3.5. а) y
2x 2
б) y x 2 e x
3.6. а) y
1
4
x 1
x 1
3.16. а) y
x
б) y x 4e 2 x
2
4x 3 5
x
x 1
б) y ln
x2
x2 1
3.18. а) y 2
x 2
2 x 1
б) y xe
3.17. а) y
3.19. а) y
x3
x2 1
1
б) y e 2 x
4x 3 5
3.20. а) y
x
1
б) y e 2 x
x3 4
3.21. а) y 2
x
x
1
x3
x3 4
3.22. а) y 2
x
x
б) y 3 3 ln
x4
x2
3.23. а) y
x 12
x
б) y 3 3 ln
x4
8x 1
3.24. а) y
x 12
x
2
б) y ln
x2
8x 1
3.25. а) y
x 12
x
1
x 1
4x
3.7 а) y
4 x2
x
б) y ln
x 1
12 x
3.8. а) y
9 x2
x6
1
б) y ln
x
б) y 2 ln
x 1
x
3.9. а) y
б) y xe x
2
2
3.10. а) y
б) y
б) y 3 ln
4x 2
3 x2
б) y 2 x 1e 21 x
e x 3
x3
1
2
x 2x
б) y x 4e 2 x
3.11. а) y
x3
3.12.а) y 2
x 2x 3
б) y ln( 1 x 2 )
x2 1
x2 2
б) y ln x 2 2 x 2
3.13. а) y
3.14.а) y
б) y ln
x3 8
2x 2
x
x 1
3.15. а) y
б) y xe2 x 1
1 2x
x x2
2
x3 4
3.26. а) y 2
x
x
2
б) y ln
x2
2
4x 1
3.27. а) y 2
x 2x 4
x
2
б) y ln
x2
x3 4
3.28. а) y 2
x
x
1
б) y 2 ln
x 1
x3 4
3.29. а) y 2
x
3 x
б) y x 2e
3.30. а) y
б) y
12 x
9 x2
e x 3
x3
TSIS. 5
INTEGRAL CALCULUS
Problem 1. Find indefinite integrals
1. а)
e
sin2 x
б) arctg x dx
1
2. а)
x
dx
г)
д)
x 1
xdx
3
2
6
д)
1 x8
x3
x 3
2
3
4. а) 2 dx
cos x3tg x 1
г)
x
5. а)
г)
1 x
3
cos 3xdx
4 sin 3x
cos 2 x
x 1 dx
3
x 4 x
4
4
1
x5
3
dx
x5
arctg x dx
3
x 1 x
dx
г)
3 cos x 4 sin x
sin xdx
9. а) 3
3 2 cos x
г)
10. а)
x 1
3
3
6
2
в)
x
3
в)
x
3
в)
x 2 dx
x 3 5x 2 8x 4
в)
x
в)
( x 2 3)
x 4 5x 2 6 dx
в)
x 2 dx
x 4 81
x
dx
4x 5
x
3x 7 dx
4 x 2 4 x 16
2
dx
x 2x 2
2
x6
dx
2 x 17
б) x 2 e 3 x dx
2
x
2
x 3dx
3
x 2 2x
д)
2
8. а)
2 x 2 3x 1
x 3 1 dx
д)
д)
sin xdx
x
4x 1
dx
x 1
1
б) x arcsin dx
x
cos xdx
7. а) ( x arctgx)dx
1 x
г)
в)
3
2x 3
dx
4x 9
3x 1
dx
2x 9
x arcsin x
б)
dx
1 x2
д)
x
1 cos x
6. а)
г)
1 x dx
2
2
б) x3 x dx
dx
x
б) e x ln 1 3e x dx
4
dx
г)
sin x tgx
x 3 dx
3. а)
x
sin 2 xdx
г)
dx
8
в)
x 1 dx
x2
4 ln x dx
x
dx
г)
2 sin x cos x 2
6x 7
dx
4 x 13
б) x ln 1 x 2 dx
x
д)
2
x
2
dx
4x 8
б) x sin x cos xdx
д)
б)
x
x
dx
x 1
2
sin 4 xdx
2
д) 2 6 x 1 dx
x 8 x 25
б) x ln 2 xdx
д) 2 5 x 2 dx
x 2 x 10
( x 2 x 1)dx
в) 4
x 2x 2 3
в)
x
x
4
6 dx
6x 2 8
3
г)
2
5
dx
3
x
б) x 2 e 3 x dx
в)
sin x
4x
x
12. а)
г)
cos xdx
11. а)
д)
ln xdx
1 ln x
x 1dx
x 1
2
x
2
x
4
5x
б)
x 4x 1
2
dx
x3
dx
3x 5
б) arctg dx
д)
2x 1
2x 1
dx
2x 1
x ln xdx
в)
2x
2
x5
dx
2x 3
2
x 3 dx
13. а)
x 1
xdx
г) 3
x 3x 2
14. а)
dx
cos 2 x2tgx 1
г)
x
x
4
15. а)
3 dx
5x 2 6
sin xdx
2
cos 2 x
x 2 dx
г) 4
x 16
e 2 x dx
16. а) x x
e e
x 3dx
г) 3 2
x x 2x
17. а)
3
г)
3
x 1 dx
5
x
x 2dx
x
4
4x
2
18. а)
в)
2x
в)
в)
3 4x x
в)
в)
x
в)
4x 2 4x 1
2
dx
2 x 17
1
б) x arcsin dx
x
д)
x
( x 5)dx
2
2x 2
2
x 4dx
x2 x 2
д) 2 2 x 3 dx
x 8 x 25
x
2
б) x3 dx
д)
д)
б)
x 1
2
dx
3x 6
x x 2 dx
x ln( x 1)dx
2
2
2x 3
8 2x x 2
2x
x 4 x 8 dx
x ln xdx
2
2
2
x2
dx
x 1
x cos 2 x
г)
x3
dx
2x 6
б) x 2 e 5 x dx
x
sin 2 xdx
д)
б)
6
x 1
( x 3)dx
в)
д)
2 x 3x 12
dx
x3 x 2 6x
2
19. а)
x arctgx
1 x 2 dx
x 4 dx
г) 4
x 6x 2 8
x
2
2x 3
dx
4x 9
б) e x sin 2 xdx
д)
x
2
x
dx
4x 5
8x 3
5 2x 2x 2
dx
20. а)
e 2 x dx
ex 1
6x 4 1
г) 3
dx
2x x 1
г)
4
x 44 x 16
dx
22. а)
2 x dx
x5
x 2 dx
г) 4
x 4x 2
x 7 dx
2
4 x 13
dx
x 4 x 13
x 2 dx
в)
2 x
в)
sin x cos x
3 3
д)
dx
x 2 6 x 16
б)
e
2
x
2
dx
3 sin x 4 cos x
2
д)
б) 7 cos t sin tdx
x 3 64
3
в)
4
21. а)
б) sin ln x dx
2 x 2 8 x 5
( x 2)dx
cos xdx
6
23. а)
д)
2x
б) arcsin 3xdx
4 ctg 2
cos 2
г)
д)
2 x 3x 12
dx
x3 x 2 6x
1 x 2 dx
24. а)
x 1 x 2
dx
г)
5 3 cos x
2
25. а)
г)
д)
z 2 dz
г)
sin2 x
x2
2
x
2
б)
x ln xdx
28. а)
2 6
dx
x 3
cos 3xdx
4 sin 3x
2
x
в)
2 x 2 3x 1
x 3 1 dx
в)
cos x3tgx 1
в)
x
x2
dx
4x 2
4
dx
3x 4
б) arctg x dx
xdx
x3
в)
3 4x 2
dx
8x 9
д)
2x
4 x
3
д)
dx
sin x tgx
dx
в)
2
б) x e dx
4 5z 6
9 7x 4
dx
4x 1
sin 2 xdx
27. а)
г)
2
xdx
в)
x 2dx
x 2x 5
3
26. а)
e
3x
б) 3 5 x e 3 x dx
1 ex
dx
1 ex
dx
3x 5
2
2
д)
x
2
dx
2
dx
x2
б) x3 x dx
dx
8
3
г)
x2 1 x
3
1 x
dx
x arctgx
1 x 2 dx
29. а)
д)
4x
2
dx
10 x 24
б) sin(ln x)dx
в)
x
2
x2
dx
x 1
д)
x 2 dx
г) 4
x 16
4x
30. а)
2
dx
5x 2
x
2
x 2 dx
в) 4
x 81
б) x3 dx
dx
cos x2tgx 1
dx
г)
2
cos x 2 sin x 2
д)
x
2
dx
x 1
Problem 2. Calculate definite integrals
16
sin x
1. а)
dx
x
0
б)
0
2
x
2
б) x 2 cosxdx
cos xdx
0
0
2
1
1dx
17. а) x
x
1
cos xdx
2. а)
1 sin x
6
2
1
sin xdx
б)
2
0 1 cos x
4
x 1dx
3. а)
x
0
б)
2
e
cos 2 x
sin 2 xdx
0
б)
2
3
sin
4 dx
2
4
4
18. а)
e
б) x 2 ln xdx
1
0
4 6 x dx
4. а)
2
2
19 а)
dx
4 5x
5
1
5
5x
dx
5. а) sin
4
sin xdx
б)
2
1 cos x
2
sin 2 xdx
3
0 cos 2 x
б)
2
8
20. а)
0
6
0
б) x 2 sin xdx
4
ctgxdx
3
б)
dx
4 3x 2
1
4
2
cosdx
2
3six 2
16. а)
cos dx
2 sin x
4
6. а) 12
0
x 3dx
21. а)
4 2
2 1 x
2
2
sin x
cos xdx
4
3
xdx
б) 2
sin x
4
3
3
4
2
xdx
1
3
8
б) x 7dx
1
dx
4 3x
7
arcsin xdx
0
e
9. а)
3
sin xdx
5 3
б)
5
24. а)
3
3
8
x 2 dx
10. а) 2 x
2
x
0
x ln xdx
1
11. а)
2
8 xdx
Sin 2 x 2
0
2
2 xdx
0
6
б)
3
16
14. а)
x 1cos 2 xdx
28. а) sin
5x
dx
4
1
dx
4 3x 2
x
0
0
4
13. а) tgxdx
2 3
x 1dx
2
x ln xdx
27. а)
4
x 1cos 2 xdx
4
sin x
dx
x
0
1
4
16
26. а)
3
Sin5x cos7 xdx
12. а)
2
б) x cos 2 xdx
2
4
б)
x dx
1
x x dx
4
0
x
3
б) x 2 e x dx
0
29. а)
2
cos
2
xdx
0
1
б) x 2 e x dx
0
4
cos
25. а)
e2
1 4x
б)
dx
ctgx sin 2 x
arctgxdx
3
2
0
б)
e2
б)
4
1
4
б)
cos2 xdx
dx
25 x 2
6
dx
9 25 x 2
5
3
cos x
sin 2 x
0
2
4
7
23. а)
5
б)
2
dx
3 4x2
0,5
б)
2
б)
7
0
x
22. а) e 3 dx
0
8. а)
9dx
9 16 x 2
0
2
cos
7. а)
4
б)
б)
x
0
2
cosxdx
б)
2
cos xdx
15. а) 3
e cos x sin xdx
30.
а)
2
sin x
0
3
2
2
e
x dx
б)
б) x 2 ln xdx
3
0 1 x
1
Problem 3.
1. Find the area bounded by the curve y sin x , the x-axis, and the lines x 0 and
x 2 .
x2 y2
2. Find the area of the ellipse 2 2 1 .
a
b
3. Find the area bounded by the curve x a cos3 t , y b sin 3 t for 0 t 2 .
4. Find the area of the lemniscate r 2 a 2 cos 2 .
2
5. Find the length of the cissoid r 2a tan sin from 0 to
.
4
6. Find the area bounded by the parabola y 2 2 x 4 and the straight line
x y 2.
7. Find the area bounded by the ellipse x 2 4 y 2 6 x 8 y 9 0 .
8. Find the area of the ellipse whose parametric equations are x a cos t and
y b sin t .
9. Find the parabola y 2 4 x from (0,0) to (-4,4).
10. Find the area bounded by the parabola y x 2 1 , the x-axis, and the lines x 1
and x 4 .
11. Find the bounded area between the curves y x 3 6 x 2 8x and y x 3 4 x .
12. Find the area of an arch of the cycloid x a(t sin t ) and y a(1 cost )
where 0 t 2 .
13. Find the area of the cardioid r a(1 cos ) .
14. Find the area of one loop of the curve r a cos 3 .
15. Find the area between the curve (2a x) y 2 x 3 where a 0 and its
asymptote.
16. Sketch the graph (2 x) y 2 x 2 (2 x) and find the area of the loop.
17. Sketch the graph (2 x) y 2 x 2 (2 x) and find the area between the curve
and its asymptote.
2x
18. Find the area in the first quadrant between the graph of y
, the
2
2
2 3
x a
coordinate axes, and the vertical line x 3a .
19. Find the circumference of a circle of radius R.
2
3
2
3
2
3
20. Find the entire length of the hypocycloid x y a .
21. Find the volume of the solid generated by revolving about the x-axis the area
3
2
bounded by the curve y x and the lines x=0 and y=8.
22. Find the volume of the solid generated by revolving about the line y=8 the area
3
bounded by the curve y x 2 and the lines x=0 and y=8.
23. The area bounded by the curve xy=1 and the lines x=1, x=3 and y=0 is evolved
about the x-axis. Find the volume generated.
24. The smaller segment cut from the circle x 2 y 2 4 by the line y=1 is revolved
about that line. Find the volume generated.
25. Find the volume generated by revolving about the line y=x the area bounded by
the line y x and the parabola y x 2 .
26. The area bounded by the hyperbola 16 x 2 9 y 2 144 and the line x=6 is
revolved about the y-axis. Find the volume generated.
27. Find the volume generated by revolving about the line x=2, the area bounded
by the parabola y 2x 2 and the line 2 x y 4 0 .
28. Find the area of the surface generated by revolving the curve r 1 cos
about the initial line.
29. Find the area of the surface generated by revolving the curve r 2 sin
about the line
.
2
30. Find the area of the surface generated by revolving an arch of the cycloid
x a(t sin t ) , y a(1 cost ) where 0 t 2 about the x-axis.
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