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Ministry of Public Health and Social Development of the
Russian Federation
Stavropol State Medical Academy
Department of medical and biological physics
I.I. Markov, O.V. Vecher, E.I. Khrynina
Practical Course on Elements of
Higher Mathematics and MedicalBiological Physics
Part I
Stavropol
2006
УДК
51; 53; 57.
ББК 22.11 я 73
П-69
Марков И.И., Вечер О.В., Хрынина Е.И.
П – 69. ПРАКТИЧЕСКИЙ КУРС ПО ВЫСШЕЙ МАТЕМАТИКЕ И
МЕДИЦИНСКОЙ И БИОЛОГИЧЕСКОЙ ФИЗИКЕ. ЧАСТЬ I. Учебное
пособие
для
студентов
английского
отделения
лечебного
и
стоматологического факультетов (на английском языке). – Ставрополь.
Изд-во СтГМА. – 2006. – 72 стр. ISBN
Авторы:
Марков Иван Иванович – д.т.н., профессор, заведующий кафедрой
медицинской и биологической физики Ставропольской государственной
медицинской академии.
Вечер Ольга Владимировна – старший преподаватель кафедры
медицинской и биологической физики Ставропольской государственной
медицинской академии.
Хрынина Елена Игоревна – старший преподаватель кафедры
медицинской и биологической физики Ставропольской государственной
медицинской академии.
Учебное пособие предназначено для иностранных студентов, изучающих
медицину на английском языке. Оно включает необходимый теоретический и
практический материал по высшей математике, а также по медицинской и
биологической физике.
Рецензенты:
Семенчин Евгений Андреевич – доктор физико-математических наук,
профессор Ставропольского государственного университета
Знаменская Стояна Васильевна – кандидат педагогических наук,
доцент кафедры иностранных языков Ставропольской государственной
медицинской академии
Чернобривая Татьяна Геннадьевна – старший преподаватель кафедры
иностранных языков Ставропольской государственной медицинской академии
Рекомендовано к изданию Советом Естественнонаучных дисциплин
Ставропольской государственной медицинской академии.
ISBN
© Ставропольская государственная
медицинская академия, 2006
2
The differential calculus (Part I)
TOPIC: Definition of the derivative. The derivatives of elementary functions.
The rules of differentiation. The definition of the elder derivatives.
The geometric and physical meaning of the derivative
Definition of the derivative
Studying various physical, biological, chemical processes and phenomena a
task of rate finding of these processes appears. The solving of this task lends to the
notion of the derivate.
The process of finding the derivative is called differentiation, and the branch of
calculus that deals with derivatives and their applications is called the differential
calculus.
Let the function y  f x is defined on a closed interval X. We take an arbitrary
point x0  X and assign an arbitrary increment x to the argument x at the point x 0 ,
the increment being such that the point x0  x also belongs to X . The function will
get an increment: y  f x0  x   f x0 .
Definition. The derivative of the function y  f x at the point x 0 is called the
limit of the ratio of the increment of the function at this point to the increment of the
argument when x  0 .
There are many symbols which are used for the marking of derivative of the
function y  f x : y x/ or y / or f / x  or
dy
.
dx
Thus, by definition: y / x0   f / x0   lim
x 0
f x0  x   f x0 
y
 lim
.
x x0
x
(1)
The definition of a derivative is a method of its calculation.
Example 1. Find the derivative of the function f x   x 2 at the point x  x0 .
Solution.
1. Assigning an increment x to the argument x at the point x 0 , we find the
corresponding increment of the function:
3
y  f x0  x  f x0   x0  x  x02  x02  2 x0 x  x  x02  2 x0 x  x
2
2
2
2 x0  x x
y 2 x0 x  x 


 2 x0  x
x
x
x
2
2. We form a ratio:
3. Find the limit of the ratio when x  0 :
y
lim x  lim 2x
x 0
x 0
0
 x   lim 2 x0  lim x  2 x0 .
x 0
x 0
Consequently, the derivative of the function f x   x 2 at the point x 0 is equal to
the number 2x0 . In symbols, we have accepted, we can write this as f / x0   2x0 .
By analogy the derivatives of all elementary functions were found. For finding
derivatives of elementary functions we can use the table of derivatives of elementary
functions.
The derivatives of elementary functions
( x n ) 
d n
( x )  n  x n 1
dx
(tan x) 
d
1
(tan x) 
dx
cos 2 x
(a x ) 
d x
(a )  a x  ln a
dx
(cot x) 
d
1
(cot x)   2
dx
sin x
(e x )  
d x
(e )  e x
dx
(arccos x) 
d
1
(arccos x)  
dx
1 x2
(arcsin x) 
d
1
(arcsin x) 
dx
1 x2
d
1
(arctan x) 
dx
1 x2
(ln x) 
d
1
(ln x) 
dx
x
(cos x) 
d
(cos x)   sin x
dx
(arctan x) 
(sin x) 
d
(sin x)  cos x
dx
(arc cot x) 
d
1
(arc cot x)  
dx
1 x2
We can use rules of differentiation for finding derivatives of functions
composed of some elementary functions.
The rules of differentiation
1. We can put the constant factor before the sign of the derivative, i.e.
y /  Сf x   C  f x   Cf / x  , where C – an arbitrary constant
/
/
2. The derivative of constant value is equal to zero, i.e.
4
d
C  0 or
dx
C /
0
If the functions U  U x and V  V x are differentiable at the point x , then the
sum, the difference, the product and the quotient of these functions are also
differentiable at this point, and the following formulas hold true.
3. The derivative of algebraic sum of functions U  U x and V  V x is equal
to algebraic sum of derivatives of these functions:
d
U x   V x   dU x   dV x 
dx
dx
dx
or U x   V x /  U / x   V / x 
4. The derivatives of product of functions U  U x and V  V x is equal to sum
products:
d
U x   V x   V x  dU x   U x  dV x 
dx
dx
dx
or U x   V x /  U / x   V ( x)  V / x   U ( x)
5. The derivative of quotient of U  U x and V  V x is found by the following
expression:
d  U x  


dx  V  x  
V x 
dU x 
dV  x 
 U x 
dx
dx
2
V x 
U
U /V  UV /
or   
(only if V x   0 ).
V2
V 
/
The geometric meaning of the derivative
Assume that the function y  f x is defined on open interval a, b and the
point N of the graph of the function corresponds to the value of the argument x 0 and
the point R to the value x0  x . We draw a straight line through the points N and R
and it is a secant. Then we denote the angle between the secant and the x -axis by
 x . This angle evidently depends on x .
If there is a limit
lim  x   
x  0
0
, then the straight line with the slope k  tan  0 ,
which passes through the point N x0 , f x0  is called the limiting position of the
secant NR when x  0 (or as N  R ).
Definition. The tangent S to the graph of the function y  f x at the point N is
the limiting position of the secant NR when x  0 , or what is the same , when
N  R.
5
y
S
R
∆y
N
∆x
φ0
φ(∆x)
a
x0
P
x0+∆x
x
b
Fig. 1
It follows the definition that for a tangent to exist it is sufficient that the limit
lim  x   
x  0
0
exists, the limit  0 being equal to the angle of inclination of the
tangent to the x-axis.
We shall prove that if the function y  f x has a derivative at the point x 0 , then
there is a tangent to the graph of the function y  f x at the point N x0 , f x0  , the
slope of the tangent (i.e. the tangent of the angle of its inclination to the x-axis) being
equal to the derivative f / x0  .
Indeed, from the triangle NRP we get: tan  x  
y f x0  x   f x0 
.

x
x
The physical meaning of the derivative
If the some physical value y it’s a function of the time t , then derivative
dy
- is
dt
a rate of changing value y . In total case, the physical meaning of the derivative
dy
dx
is
the speed with which the values of the function y (x) change when x changes.
For example:
1. If the distance which a body travels in time t is given the value at t of a
function f, S=f(x), then the instantaneous velocity of the body at time t is

dS
S
 St/  lim
.
t  0 t
dt
6
2. If the velocity  which a body moves in time t is given by the value at t of a
function f, υ=f(x), then the instantaneous acceleration of the body at time t is
a
d

 t/  lim

t

0
dt
t
3. If the angle of rotation  which a body travels in time t is given by the value
at t of a function f,   f (x) , then the angular velocity of the body at time t is:

d

 t/  lim
t  0 t
dt
4. If the quantity of heat Q which a body gives out or takes in temperature T is
given by the value at T as a function f, Q=f(T), then the heat capacity of the body at
temperature T is:
С
dQ
Q
 QT/  lim
T  0 T
dT
5. If the electric charge q which an electrical charged body and moves in time t
is given by the value at t of a function f, q=f(t), then the electric current is
i
dq
q
 qt/  lim
dt
t 0 t
The definition of the second derivative
The derivative from a given function y=f(x) is also the function at the argument
x. For example: If y=x3, then y/=3x2.
The derivative of the derivative of a function is called the second derivative. If
the function y=f(x), then we denote the second derivative y //  f // ( x) 
d2y
, read “f
dx 2
second” or “f double prime”.
It follows from the definition of derivative that the second derivative is a
function. In general, the second derivative will itself have a derivative, called the
third derivative of the original function, and this process of finding successive
derivatives is one which, in many cases, can go on forever.
As the preceding analysis indicates, negative values of the second derivative
correspond to regions in which the slope is decreasing and the graph is therefore
concave down, and positive values of the second derivative correspond to regions in
7
which the slope is increasing and the graph is concave up. A point where the
direction of curvature changes is called a point of inflection of the curve and is
evidently a point at which sign of the second derivative changes.
The physical sense of the second derivative
If the distance s which a body travels in time t is given the value at t of a
function f, S=f(t), then the instantaneous acceleration of the body at time is:
a
d d 2 S
 2  St//
dt
dt
The instantaneous acceleration of a body, which moves irregularly, is the
second derivative.
The definition of the elder derivatives
The derivative of the second derivative is called the third derivative. We
d3y
denoted the third derivative y  f ( x)  3 “f third” or “f three prime”.
dx
///
///
The derivative of the third derivative is called the fourth order derivative. We
denoted the fourth-order derivative y IV  y ///  
/
d4y
and so on.
dx 4
All derivatives, starting from the second, are called the elder derivatives.
5
/
4
//
3
///
2
For example: y  x , then y  5x , then y  20x , then y  60x , then
y IV  120 x .
The rule of differentiating a composite function
If the function x   t  has a derivative at the point t 0 and the function y  f x
has a derivative at the corresponding point x0   t 0  , then the composite function
f  t  has a derivative at the point t 0 and the following formula holds true:
y / t 0   f
/
x0    / t 0  or
d
 y   df  dx .
dx
dx dt
This formula is called “Chain rule”.
Example 2. Find the derivative of a composite function y  ln cos x .
8
Solution. In this case y /  ln cos x / 
1
1
/
 sin x    tan x .
 cos x  
cos x
cos x
Exercises
I. Differentiate:
1. y  3x 4  5x 3  2 x 2  x ;
2. y  cos 2 x ;
16. y  (2 x  5) 3 ;
17. y  e x  cos x ;
3. y  arctan 3x ;
18. s  (2  4 t 3  t 3 )(  t ) ;
4. y  (4 x 3  2 x 2  5x)  ( x 2  7 x) ;
19. y  5 x ;
2
t
2x  3
.
3x  7
21. y  e x  ln x  3 ;
5. y 
x
;
x 1
6. y  cos x  (3 sin x  2) ;
20. y 
7. y  esin x ;
22. y  ln(sin x) ;
sin x  x
;
ln x
1 x
9. y 
;
x 1
1
10. y  x x 
;
3 x
11. y  x  ln x ;
8. y 
23. y  ecos 3 x ;
24. y  7 x 2  5 cos3 x ;
25. y  e3x  7 x5 ;
2
26. y  ln 3 2 x ;
27. y  (1  4 x3 )(1  2 x 2 ) ;
12. y  ln(cos x) ;
13. y  sin 3x  cos 4 x ;
28. y 
12 cos x
;
1  sin x
14. y  cosx4  2 x3  3x  ;
29. y 
3x 4 3 x 2
;
3
x
15. y 
1
1 5 1
 2  ;
3
x
x
x 4
30. y  2 cos(ln 3x) .
II. Solve these tasks with the help of the derivative:
1. The distance an object travels in time t is S , find its velocity and
acceleration, where S is given as:
a) S  9t  t 3 and t  2 s.
b) S  cos
t
3
and t  1s.
2. Find the time moment when the acceleration is equal to zero, if the distance
1
6
are object travels in time t is S , where S   t 3  3t 2  5 . Find its velocity.
3. An object which mass is 6 kg travels in time t , where S  1  ln( t  1)  (t  1)3 ,
find its kinetic energy, when t  1s after the beginning of motion.
9
The differential calculus (Part II)
TOPIC: The differential. Geometric meaning of the differential. Differentials
of higher order. Particular derivatives. Whole differential.
The definition of the differential
Assume that the function y  f x is differentiable at the point x 0 , i.e. its
increment y at this point can be written as the sum of two terms:
y  Ax  x ,
where
lim x  0 . If
x  0 the term Ax is a very small value of the same order as
x 0
x (for A  0 ). It is the linear relative to x . As x  0 , the term x is very small
 x

 0 .
value of a higher order than x  lim

x  0
x

Thus, for A  0 , the first term is the principal part of the increment of the
function y  f x .
Definition. The differential of the function y  f x at the point x 0 is the
principal, linear relative to x , part of the increment of the function at this point:
dy  Ax
(1)
If A  0 , then Ax  0 , and therefore the term Ax is no longer the principal
part of the increment y since the term x is in general, different from zero.
However, by definition, in this case as well as the differential of the function at the
point x 0 equals to Ax , i.e. dy  0 .
Taking into account that A  f / x0  , we can write formula (1) in the form
dy  f
/
x0 x
(2)
Let f x  x . Then, by formula (2),
x  x   x0 
x 


/
dy  dx  x0  x   lim 0
x   lim x  1  x  x
x
 x0 x 
 x0

The differential of the independent variable x is the increment dx  x of this
variable. Now, relation (2) is assumed in the form
10
dy  f
/
x0 dx
(3)
Note that using relation (3), we can calculate the derivative f / x0  as the ratio
of the differential dy of the function to the differential dx of the independent
variable, i.e. f / x0  
dy
.
dx
Example 1. Find differential of the function y  sin 3x .
Solution: Firstly we must find the derivative of this function: y /  3 cos 3x . And
differential is equal to: dy  3 cos 3xdx .
The differential of a function has a geometric meaning.
Assume that a point M on the graph of the function y  f x corresponds to the
value of the argument x 0 , a point P, to the value of the argument x0  x , a straight
line MS is a tangent of the graph of y  f x at the point M, and the angle between the
tangent and the x –axis. Assume, furthermore, that MN // OX , PN // OY and Q is the
point of intersection of the tangent MS and the straight line PN. Then the increment
y of the function is equal to the value of the line segment NP. At the same time,
from the right triangle MNQ we get NQ  tan x  f / x0 x  dy , i.e. the differential
dy of the function is equal to the value of the line segment NQ. In terms of geometry,
the values of the segments NP and NQ are different.
y
P
Q
S
∆y
y  f x 
M
N
∆x
α
x
x0+∆x b
x0
Fig. 1.
11
Thus the differential dy of the function y  f x at the point x 0 is equal to the
increment of the “ordinate of the tangent” MS to the graph of this function at the
point M x0 , f x0  , and the increment y of the function is the increment of the
“ordinate of the function” y  f x it self at the point x 0 corresponding to the
increment x of the argument.
Differentials of higher orders
Consider differentials of higher orders. For the sake of convenience, we shall
use the designations dy and dx .
Let the function f x  is differentiable at every point x of a certain closed
interval. Then its differential dy  f / x dx , which we shall call the first-order
differential, is the function of two variables of the argument x and of its differential
dx . Let the function f / x  , in its turn, is differentiable at a point x . We consider dx in
the expression for dy as a constant factor. Then the function dy is the function of
only the argument x and its differential at the point x has the form:

 

d dy   d f / x dx  f / x dx dx  f
/
//
x dxdx
(4)
The differential d dy  designated d 2 y and called the second-order differential.
We can write formula (4) as:
d2y  f
//
x dx 2
(5)
In its turn, the differential d d 2 y  of the differential d 2 y is the third-order
differential of the function f x  and is designated as d 3 y and so on. The differential


d d n 1 y of the differential d n 1 y is the n-th order differential (or n-th differential) of
the function f x  and is designated as d n y .
We have proved that the formula
d n y  y n  x dx  , n  1, 2, 3,...
n
is valid for the n-th order differential of the function.
12
The differential calculus of functions of some variables. The partial
derivative. The total differential
Most of processes happening in nature are submitted to laws expressing
dependence between some variables.
For instance:
1) the equation of a gas state is the dependence between three values: pressure,
volume and temperature:
pV
 c;
T
2) the rate of bacteria growth depends on concentration of a medium, and
temperature, and on illumination;
3) any physiological characteristic of an organism (weight, temperature, height,
pressure) is a function of many variables.
Function of some variables is called functional dependence between some
variables number of variables is more then one.
The function of two variables is: z=f(x, y).
The function of three variables is U=f(x, y, z).
The variable U is called the function of three variables x, y, z, if for every three
variable x, y, z is the only value of U.
The function of n variables x, y, z, k, … t is U=f(x, y, z, k, … , t)
The particular derivatives. The definition of the particular derivative
The particular derivative of a function of some variables on one of these
variables is called the limit of a corresponding particular increment of function to an
increment of given function, when the last one approaches to zero.
The symbol of the particular derivative is:
df ( x )
f ( x )
instead
.
dx
x
For example: z=f(x, y, t) is the function of some variables, then z x/ 
the particular derivative on x and y=const, t=const.
z y/ 
f ( x, y, t )
is the particular derivative on y and x=const, t=const.
y
13
f ( x, y, t )
is
x
zt/ 
f ( x, y, t )
is the particular derivative on t and x=const, t=const.
t
The whole derivative is called the sum of particular derivatives on all
arguments:
z x , y , t  z x/  z y/  zt/ 
f ( x, y, t ) f ( x, y, t ) f ( x, y, t )


x
y
t
Finding of particular derivative of simple functions is made in accordance with
the known rules of the differentiation of function of one variable. In finding the
particular derivative on any argument all other argument are considered to be
constant.
The total differential of a function.
The whole differential of a function of two variables is equal to the sum of
particular differentials: dz 
z
z
dx  dy  z x/ dx  z y/ dy . For finding the whole
x
y
differential it is necessary to find the whole derivatives, and then each particular
derivative is multiplied by endlessly small increment of its argument.
Applications of the differential for the estimation of errors
The true value of any physical size can not be taken as a result of
measurements. So experimental results obtained due to the measurements usually
contain errors. There are absolute errors and relative errors.
1) The absolute error is called the absolute value of the difference between the
real value and the approximate value of a function: y  f x  x  f x .
2) The relative error is called the absolute value of the relation of the absolute
error to the value of a measured size:  
y
.
y
An important application of the differential is to approximation of irrational
roots of equations. This process, which is known as Newton’s method, is best
explained through a specific example.
14
EXCERCISES
I. Find the differential of the functions:
1. y  e3 x  5 ;
2
2. y  7 cos 5 x ;
3. y  ln 3 1  x 2 ;
4. y  cos x sin x ;
5. y 
5 1
 ln x ;
x2 4
6. y  ecos 3 x ;
7. y  3 4 x 2  2 ;
8. y  sin 3 x  2 cos4 x ;
2
x
9. y  1  ;
10. y  4 x 2 ln x ;
II. Find the whole differential of these functions:
1. z  ln y 3 ;
2. z  x 4 y  2 xy ;
3. z  ln 6 x3  3 y 4 ;
4. U  5x 2 cos 4 x  ln(sin y) ;
5. U  12 x 4  3 y 5  7 z 6 ;
6. z 
x 1
;
y2
7. z  3x 2 y 3  2ax  5 y 3  3 x y 3 ;
8. z 
3
 7 x3 y 5  2 y 3 ;
x2
9. U  ecos x sin y 
10. z 
1
xy ;
z2
1
1
 2.
2
x
y
15
The Integral Calculus (Part I)
TOPIC: Antiderivative. Definition of the indefinite integral. Principal
properties of the indefinite integral. Principal methods of integration
The antiderivative
Seeking a derivative of a given function is one of the most important problems
of differential calculus. Various problems of analysis and its numerous applications
in geometry, mechanics, physics and engineering lead to the converse problem: using
the given function f x  , we must find a function F x  whose derivative would be
equal to the function f x  , i.e. F / x   f x  .
Finding a function from the known derivative of that function is one of the
fundamental problems of integral calculus.
Definition. The function F x  is an antiderivative of the function f x  of an
interval X if for all values of x from this interval the equality
F / x   f x 
(1)
holds true.
The problem of finding an antiderivative, being given function f x  , does not
have a unique solution. Indeed, if F x  is an antiderivative of f x  , i.e. F / x   f x  ,
then the function F x  C , where C is an arbitrary constant, is also antiderivative of
f x  since F  x   C   f  x  for any number C.
/
We shall now show that the set of functions F x  C , where F x  is an
antiderivative of the function f x  and C is an arbitrary constant, exhaust all the
antiderivative of the function f x  .
It the F x  is an antiderivative of the function f x  on an interval X , then any
other antiderivative of f x  on this interval can be represented in the form F x  C ,
where C is an arbitrary constant.
For example, if we have function
f x   2 x ,
equals F x   x 2  C . Consequently, F / x   x 2  C   2 x .
/
16
then its antiderivative
Definition of the indefinite integral
Definition. If the function F x  is an antiderivative of the function f x  on an
interval X , then the set of functions F x  C , where C is an arbitrary constant, is an
indefinite integral of the function f x  on this interval and is designated as:
 f x dx  F x   C
(2)
In the case the function f x  is the integrand, f xdx – the element of
integration or underintegral expression, x – is the integration variable, C – an
additive constant, which includes the totality of all antiderivatives of the given
function. For finding C it is necessary to give initial conditions.
Thus the symbol
 f x dx
denotes the collection of all antiderivatives of the
function f x  . But sometimes we shall understand it to be any elements of that
collection, i.e. one of the antiderivatives.
Finding a function being given its derivative, or, that is the same, seeking an
indefinite integral being given an integrand, is the integration of that function.
Integration is the inverse of differentiation. To verify whether integration was carried
out correctly, it is sufficient to differentiate the result and obtain the integrand.
The geometric meaning of the indefinite integral
Geometrically the indefinite integral is presented by a family of functions,
which equations differ from each other by additive constant only. The graph of the
functions getting result of integration is called integral curves.
Each integral gives a family of integral curve. Integral curves of one family
have the same shape and are displaced relatively each other along a vertical
determined by the value C. For example,  2 xdx  x 2  C (see graph. 1).
y=x2+C
y=x2
y
y=x2-C
x
Fig. 1
17
Principal properties of the indefinite integral
1. The derivative of an indefinite integral is equal to the integrand, i.e.
 f x dx   f x 
/
2. The differential of an indefinite integral is equal to the element of integration,
i.e.
d  f  x dx  f  x dx
3. An indefinite integral of the differential of a function is equals to the sum of
the function and the arbitrary constant, i.e.
 dF x   F x   C
4. We can put the constant factor before the sign of the integral, i.e. k  const  0 ,
then
 kf x dx  k  f x dx
5. An indefinite integral of the algebraic sum of two functions is equal to the
algebraic sum of the integrals of those function taken separately, i.e.
  f x   g x dx   f x dx   g x dx
The table of the basic integrals
dx
1)
 dx  x  C ,
8)
 cos
2)
x n 1
 x dx  n  1  C, for n  1
9)
 sin
3)

4)
e
5)
x
 a dx 
6)
 sin xdx   cos x  C ,
7)
 cos xdx  sin x  C ,
n
dx
 ln x  C , x  0,
x
x
x
  sec 2 xdx  tan x  C ,
x
  cos ec 2 x   cot x  C ,
2
dx
2
10)  tan xdx   ln cos x  C
dx  e x  C ,
11)  cot xdx  ln sin x  C ,
ax
 C,
ln a
18
dx
12)

13)
1 x
1 x
dx
2
2
 arcsin x  C ,
 arctan x  C.
Principal methods of integration
1. Direct integration
Calculation of integrals by a direct use of the table of the simplest integrals and
the principal properties of indefinite integrals is called a direct integration.
Example 1. Integrate:
 5x
2

 4 dx .
Solution. Using properties 5 and 4, we can write


x3
5 x  4 dx  5 x dx  4 dx  5
 4x  C .
3
2
2
2. The method of substitution
In many cases the introduction of a new variable marks it possible to reduce
seeking a given integral to seeking a tabular integral, i.e. to pass to a direct
integration. This method is known as the method of substitution, or the method of
change variable. It’s based of the following formula:

x   t 
f x dx 
dx   t dt
/
  f  t  / t dt
(3)
holds true on the set T. At the end it is necessary to return to the first variable
(or x variable).
Example 2. Integrate:  sin 3xdx .
Solution:
We can find antiderivative in this case using method of substitution. Let t  3x ,
then dt  3dx and dx 
dt
. Then we can write
3
1

1
1
1
 sin 3xdx   sin t  3 dt   3  sin tdt   3 cos t  C   3 cos 3x  C
Example 3. Integrate:  sin 2 x cos xdx
Solution: Let t  sin x , then dt  cos xdx and dx 
and instead dx expression
dt
we get:
cos x
19
dt
. Put instead sin x value t
cos x
2
2
 sin x cos xdx   t cos x
dt
t3
1
  t 2 dt   C  sin 3 x  C
cos x
3
3
3. The method of integration by parts
This method is based on the use of the formula for the differentiation of the
product of two functions.
Assume that the function U x  and V x are defined and differentiable on an
interval X and let the function U / x V x  have an antiderivative on this integral. Then,
on the interval X, the function U x V / x  also has antiderivative and there holds a
formula:
 U x V x dx  U x V x    V x U x dx
/
/
(4)
Formula (4) is known as the formula for integration by parts in an indefinite
integral. Since V / x dx  dV and U / x dx  dU , it can be rewritten as
 UdV  UV   VdU
(5)
This formula makes it possible to reduce the calculation of
 UdV
to the
calculation of the integral VdU , which may prove to be simpler.
This method is usually used, if integrand expression is a product of algebraic
function on one of the following function: sin x, cos x, ln x, e x , arcsin x, arctan x .
Example 4. Integrate:  x  e x dx .
Solution: In this case integral expression is a product of algebraic function and
the function e x . Using formula (5) we can write:
x
 xe dx 
Ux
dU  dx
dV  e dx
x
V e
x
 xe x   e x dx  xex  e x  C
Exercises
Perform the following integrations:
1.  5 x 2 (3  2 x3 )dx ,
16.
20
dx
 1  2x
2
,

2.
3t  2dt ,
x3
dx,
2
3.

4.
4
 2 x  5 dx ,
5.
17.
x 3  10
 x 4 dx ,
18.
 1  cos
sin xdx

 x dx,
20.

6.  sin 4d ,
21.
x
x 3  3x 2  1
dx,
x5
7.

8.
 sin
dx
2
x
,
9.  (6 x 5  5 x 4  3 x 2  2 x  1)dx ,
dx
,
x
10.
4
11.
x
dx
4
,
12.  (6 v  5v 2 )dv,
,
x
e x dx
19.
8
2
,
1  e2 x
1  sin 2 x dx ,
x 2  3dx ,
22.
 ln xdx ,
23.
 4  2x ,
dx
x3dx
24.
 x
25.
 x sin xdx ,
26.
 x ln xdx ,
27.
 x  5 cos xdx ,
4
5

3
,
13.
x 4  3x3  1
 x dx ,
28.  (6  2 x) 4 dx ,
14.

29.
 x  1e
30.
 cos
15.
5
1  2 x dx ,
x
2
dx
,
 2x  1
21
x
2
x
x
dx ,
dx .
The Integral Calculus (Part II)
TOPIC: Definition of the definite integral. Principal properties of the
definite integral. The Newton-Leibniz formula. Some applications of the definite
integral in physics and geometry. The formula for integration by parts in the definite
integral. Integration by substitution in definite integral.
In contradistinction to an indefinite integral which is called the totality of all
antiderivatives, the definite integral is a number. Various concrete problems lead to
the concept of definite integral such as: calculation of plane areas; calculation of
volumes of different bodies; calculation of distances traveled by bodies if the velocity
is a function of time; calculation of work made by a force and so on.
Definition of the definite integral
Definition. Let the function y  f (x) is defined on the closed interval a; b .
We divide a; b into n arbitrary parts by points: a  x0  x1  x2  ...  xn1  xn  b . We
designate this division as  and the points x0 , x1 ,..., xn as the points of division. In each
of the resulting subintervals xi 1 , xi  we choose an arbitrary point  i . We designate as
xi the difference xi  xi 1 which we agree to call the length of the subinterval xi 1 , xi  .
n
  f (1 )  x1  f ( 2 )  x 2  ...  f ( n )  x n   f ( i )  xi , which we
We form a sum
i 1
shall call the integral sum. The geometric meaning of the sum  is obvious: it is the
sum of the areas of rectangles with the bases x1 , x2 ,..., xn and altitudes
f (1 ), f ( 2 ),..., f ( n ) (fig. 1).
y
B
A
0
a=x0 ξ1
x1
ξ2
x2
xi+1 ξi
Fig. 1.
22
xn-2 ξn-1 xn-1 ξn
xn=b
x
x
N
R
∆
y
∆
We designate the length of the largest subinterval as  .
Definition. If there is a limit of the integral sum as   0 , then this limit is
the definite integral of the function f (x) with respect to the closed interval a; b and
is designated as follows:
b

n
f ( x)dx  lim  f ( i )xi ,
 0
a
(1)
i 1
b
where
 f ( x)dx - is the symbol of the definite integral.
a
In this case the function f (x) is said to be integrable f (x) on a; b . The
numbers a and b are the lower and the upper limit of integration respectively, f (x) is
the integrand.
Principal Properties of the Definite Integral
1.
When the integrations limits coincide
a
 f ( x)dx  0
(2)
a
2.
Permutation of the limit of integration is equal to a change of the sign
which is in front of the definite integral:
b
a
a
b
 f ( x)dx   f ( x)dx .
(3)
3. The interval of integration may be divided into some parts:
b

c
b
f ( x)dx   f ( x)dx   f ( x)dx .
a
a
(4)
c
4. A constant factor can be put before the sign of the definite integral:
b
b
a
a
 k  f ( x)dx  k  f ( x)dx .
(5)
5. The definite integral of the algebraic sum of functions f (x) and  (x ) is
equal to the algebraic sum of their definite integrals:
b
b
b
a
a
a
  f ( x)   ( x)dx   f ( x)dx    ( x)dx .
23
(6)
The Newton-Leibniz formula
It is difficult to calculate definite integrals using the method based on the
definition of the integral as a limit of an integral sum.
Newton-Leibniz formula yields a simple method of calculating the definite
integral: the definite integral of a continuous function is equal to the difference of the
values of any one of its antiderivatives calculated for the upper and the lower limit of
integration:
b
 f ( x)dx  F ( x)
b
a
 F (b)  F (a)
(7)
a
2
Example 1. Calculate  e x dx.
1
2
Solution:  e x dx  e x 12  e 2  e1  e(e  1).
1
9

Example2. Calculate
x dx .
4
Solution:
9
9

4
1
1
2
9
3
3
3
 2
x
2 2
2  2
x dx   x dx 
 x   9  4 2  
1
3 4 3
4
 3
1
2
4
9
1
2
1
2
Example 3. Calculate

0
1
2
Solution:

0
dx
1 x 2
2

1
2

1
1  x2
 arc sin x
Example 4. Calculate
Solution:
dx
1
2
0
9
3
.
 arcsin
dx
x
dx
 ln x 12  ln 2  ln 1  ln 2
x
2
Example 5. Calculate  (3x 2  1)dx
0
24
1

 arcsin 0  .
2
6

 43 
2
27  8  38
3
3
2
Solution:
 (3x
2
 1)dx  ( x 3  x) 02  (2 3  2)  (0 3  0)  6
0
Some Applications of the Definite Integral in Physics and Geometry

The area of a curvilinear trapezoid.
Assume that a figure given on the xy-plan is bounded by the closed interval
a; b of the x-axis, the straight lines
x  a and x  b and the graph of the continuous
and nonnegative function f (x) on a; b . This is a curvilinear trapezoid whose area
can be calculated by the formula:
b
b
a
a
S   f ( x)dx or S   ydx .

The volume of a body of revolution.
Let the function f (x) be continuous and nonnegative on the closed interval
a; b . Then the body which results from the rotation of a curvilinear trapezoid
bounded from above by the graph of the function f (x) about the x-axis has a volume:
b
V     f 2 ( x)dx
a

The work of a variable force.
Assume that a particle moves under the action of a force F directed along the
x-axis and having a variable value dependent on x. We have to find the work done by
the force F in displacing the particle along the x-axis from the point x  a to the point
x  b . We assume the function F(x) to be continuous on the closed interval a; b .
Work of the variable force on the interval a; b is equal to the definite integral:
b
A   F ( x)dx .
a
The Formula for Integration by Parts in the Definite Integral
If the functions u (x) and v(x) have continuous derivatives on the closed
interval a; b , then there holds a formula
25
b
b
 u  dv  u  v   v  du
b
a
a
(8)
a
This formula is known as the formula for integration by parts in the definite
integral.
e
Example 6. Calculate  ln xdx.
1
Solution: we set u  ln x, dv  dx , and then du 
e
e
1
1
that  ln xdx  x  ln x 1e   x 
dx
, v  x and from formula (8) we find
x
dx
 ( x  ln x  x) 1e  1.
x
Integration by Substitution in Definite Integral
Let f (x) be a continuous function on the closed interval a; b . Then, if
1) the function x   (t ) is differentiable on  ;   and  (t ) is continuous on  ;   ,
2) the interval a; b is the set of values of the function x   (t ) ,
3) and a   ( ) and b   (  ) , then the formula
b


f ( x)dx   f ( (t ))   (t )dt
a
(9)

holds true.
Formula (9) is known as the formula for a change of variable in a definite
integral.
Whereas when calculating an indefinite integral by a change of variable, we
must return from the new variable t to the old variable x, we may not do that when
calculating a definite integral.
e
Example 7. Calculate
ln x
dx.
x
1

Solution: we consider the substitution t  ln x , then dt 
and   ln e  1 . Using formula (9), we get
26
dx
,
x
  ln 1  0
t  ln x
1
ln x
t2
dx
.


t

dt

dx
1 x
0
2
dt 
x
e
1
0

12 0 2 1

 .
2
2 2

Example 8. Calculate  cos 3 x  sin xdx.
0
Solution: we consider the substitution t  cos x , then dt   sin xdx ,   cos 0  1
and   cos   1. Using formula (9), we get

t  cos x
1
1
t4
0 cos x  sin xdx  dt   sin xdx   1 t  dt  1t  dt  4
3
3
3
1
1
14 (1) 4 1 1
 
  0
4
4
4 4
Exercises

17.  x5 1  x 6 dx ;
0
1.  sin xdx ;
1
0
1
2.

10
ln 3 x
dx ;
18. 
x
1
1  x dx ;
0
 /2
1
3.
x3dx
0 3  x4 ;
19.
ln 2
4.
e
cos xdx
;
3
sin
x
/4


2
2x
20.  x  ln xdx ;
dx ;
1
0
 /2
5.
2
2
 cos x  sin xdx ;
21.
6.

1
2
;
22.
e

cos 2 x
 sin 2 xdx ;
/6
 /6
7.
x  1dx ;
 /4
xdx
5  4x
3
1
0
1

2x2  1
1 x dx ;
2
 sin 6 xdx ;
23.
0
x2  2x  3 x
8. 
dx ;
x
0
1
2
24.
1
1
e
e x dx
9.  x ;
e 1
0
2 xdx
 2 x
2
 1
ln 2 xdx
25. 
;
x
1
27
2
;
 /4
10.

0
2
xdx
;
cos 2 x
26.  e x  2xdx ;
1
e4 x 3
dx ;
27. 
4
1
0
2
dx
11. 
;
3
 2 1  2 x 
 x
2
12.
2

3 x 1
0 2 x dx ;
16
 2 x  3 dx ;
28.
1


cos 2 x
29.  2 dx,
 sin x
2
sin xdx
13. 
;
2
 / 2 (1  cos x)
4
 /2
14.
30.  x2 x 2  1dx ;
2
 2 sin x cos xdx ;
1
0

 3 4 x
dx ;
15.   
3 
x
1
4


2
16.
2  tan x
;
2
cos
x
0
4
31.
ex
1 x dx ;
32.  ( x  1)  sin xdx .

28
Ordinary differential equations
TOPIC: Ordinary differential equation. First order differential equations. The
general and particular solutions of differential equations
Differential equations occupy a place of their own in mathematics.
Mathematical research of various natural phenomena often leads to the solution of
these equations since the laws which a particular phenomenon obeys are themselves
written as differential equations.
Differential equations are equations in which the unknown function appears
under the sign of a derivative. The main task of the theory of differential equations is
to study functions which are solutions of this kind of equations.
Differential equations can be classified as ordinary differential equations, in
which the unknown functions are functions of one variable, and partial differential
equations, in which the unknown functions are functions of two or more variables.
The theory of partial differential equations is more complicated and is
discussed in more complete and special courses in mathematics. In this chapter we
present elements of the theory of ordinary differential equations. In what follows,
when we speak of differential equations, we mean only ordinary differential
equations.
First-Order Differential Equations
We begin studying the theory of differential equations with first-order
differential equations which are the simplest equations of this kind.
Definition of the first-order differential equation
Definition 1. An equation of the form
F (x, y, y') = 0,
(1)
where x is an independent variable, y is the sought-for function, and y' is its
derivative, is a first-order differential equation.
29
If equation (1) can be solved for y', then it assumes the form
y /  f x, y 
(2)
and is a first-order equation resolved for the derivative. We shall exclusively deal
with equations of this kind.
In some cases it is convenient to write equation (2) in the form
dy
 f  x, y  or
dx
in the form f x, y dx  dy  0 , which is a special case of the more general equation
Px, y dx  Qx, y dy  0 ,
(3)
where P (x, y) and Q (x, y) are known functions. The symmetric form (3) of the
equation is convenient in that the variables x and y are equivalent, i.e. each of them
can be considered to be a function of the other.
Here are some examples of differential equations of forms (2) and (3):
y /  xe y , y / 
y ln x
,
x
y/  x  y ,
xdx  ydy  0
Solution of an equation
Definition 2. The solution of a first-order differential equation is a
function y   x  , x  a, b , which, being substituted into the equation, turns it into an
identity.
Thus, for instance, the function y  x 3 , x   ,   is a solution of the equation
3 y  xy /  0 , i.e. being substituted into the equation, it turns it into an identity, i.e.
3x 3  3x 3  0 .
The graph of the solution of a differential equation is called an integral curve.
The general and a particular solution of an equation
We give two main definitions.
Definition 3. The general solution of equation (2) in a domain G of the xyplane is a function y   x  , which depends on x and on the arbitrary constant C if it is
a solution of equation (2) for any value of the constant C and if for any initial
30
conditions such that x0 , y0   G there is a unique value of the constant C  C0 such
that the function y   x, C  satisfies the given initial conditions  x0 , C0   y0 .
Definition 4. A particular solution of equation (2) in the domain G is a
function y   x, C0  which results from the general solution y   x, C  for a definite
value of the constant C  C0 .
In terms of geometry, the general solution y   x, C  is a family of integral
curves on the xy-plane which depends on one arbitrary constant C, and the particular
solution y   x, C0  is one integral curve of this family which passes through the
given point x0 , y0  .
Example 1. Consider an equation y' = 3x2.
Solution: this is a first-order differential equation. Using definition of the
derivative we can write this equation in the form:
dy
 3x 2 . Then dy  3x 2 dx . Using
dx
method of direct integration we can write:  dy  3 x 2 dx .
Function y  x 3  C , where C is an arbitrary constant, is the general solution of
this equation on the entire xy-plane.
In terms of geometry, this general solution is a family of cubic parabolas. For
different values of the constant C we get different solutions of this equation. For
instance, if С  0 , then y  x 3 , if C  1 , then y  x 3  1, if C  2 , then y  x 3  2 and
so on (Fig.1).
Fig.1
31
Geometric meaning of an equation
Let us be given a first-order differential equation y /  f x, y  and assume that
the function y   x  is its solution. The graph of the solution is a continuous integral
curve through every point of which we can draw a tangent. It follows from this
equation that the slope y' of the tangent to the integral curve at its every point x, y  is
equal to the value of the right-hand side of the equation f x, y  at this point. Thus the
equation y /  f x, y  establishes a relationship between the coordinates of the point
x, y  and the slope
y / of the tangent to the graph of the integral curve at the same
point. Knowing x and y, we can indicate the direction of the tangent to this integral
curve at the point (x, y).
Let us put every point (x, y) of the integral curve into correspondence with a
directed segment whose slope is f x, y . We obtain the so-called direction field of
this equation which discloses the geometric meaning of the first-order differential
equation.
Thus, from the point of view of geometry, the equation y /  f x, y  defines the
direction field on the xy-plane, and the solution of the equation is the integral curve
the direction of the tangent to which at every point is coincident with the direction of
the field at that point.
Constructing the direction field of the given differential equation on the plane,
we can approximate the integral curves.
The kinds of the first-order differential equations
The most often meeting equations of this kind are subdivided into the following
four types:
1) equations with variable separable;
2) homogeneous;
3) linear equations;
4) equations converted into the linear type by some transformations.
32
Equations with variables separable
Definition 5. An equation of the form
y /  f 1 x   f 2  y 
(4)
where f1 x and f 2  y  are continuous functions, is known as a differential equation
with variables separable.
To find the solution of equation (4), we must, as they say, separate the
dy
dx
variables in it. For that purpose, we substitute
for y / in (4), divide both sides of
the equation by f 2  y  (we assuming that f 2  y   0 ) and multiply by dx . Then equation
(4) assumes the form
dy
 f 1  x dx
f 2 y
(5)
In this equation the variable x appears only on the right-band side and the
variable y only on the left-hand side (i.e. the variables are separated).
Assuming that the function y   x  is a solution of the equation and
substituting it into (7), we get an identity. Integrating the identity, we get
 f  y   C   f x dx  C
dy
1
1
2
2
or
 f  y    f x dx  C
dy
(6)
1
2
where C  C2  C1 is an arbitrary constant.
Relation (6) implicitly defines the general solution of equation (6).
Example 2. Solve
dy

dx
xy .
1
2

1
2
Solution: As we have seen, this can be written in the form x dx  y dy  0 , and
therefore
x
1
2

1
2
dx   y dy  C ,
3
1
2 2
x  2 y 2  C,
3
33
3
1
3
2
or, somewhat more neatly, x 2  3 y 2  C / , where С /  С is also an arbitrary
constant.
Example 2. The slope of a curve at the point x, y  is given by the value of

x
. Find the equation of the family of curves which satisfy this condition and of the
y
particular solution that through 1,  3 .
Solution: The given condition is equivalent to the differential equation
dy
x
  , separating the variables and integrating, we have
dx
y
x dx  ydy  0 ,
 xdx   ydy  0 ,
1 2 1 2
x  y C,
2
2
x2  y 2  C / .
For C /  0 , this represented the family of circles with center at the origin and
radius C / ; for C /  0 , there is no corresponding curve (what does the case C /  0
represent). If a member of the family passes through 1,  3 , we have 12  (3)2  C /
and the particular solution required is therefore x 2  y 2  10
Some application of first-order differential equations
I. Problem on the radioactive decay. It has been established by experiments
that the rate of decay of a radioactive substance, i.e. the rate of change of its mass
depending on time, is directly proportional to its amount. Let us find the law of
change of the mass m of the radioactive substance with time t, assuming that at the
moment t = 0 the initial mass of the substance was m0.
Assume that at the moment t the mass of the substance is m, at the moment
t  t the mass is m  m . During the time t the mass m disintegrates. The ratio
34
m
is the average rate of decay during the time t , and
t
m
lim t
t 0

dm
is the
dt
instantaneous rate of decay at the moment t. According to the hypothesis,
dm
  km
dt
(7)
where k is the proportionality factor (we have taken the minus sign because
the mass of the substance diminishes with time and the derivative of a decreasing
function is negative). We have got a first-order differential equation from which we
must find the dependence of mass m on time t.
Solving the equation, we find that
dm
  kdt , ln m  kt  ln C
m
whence
m  Ce  kt
(8)
Formula (8) defines the mass of a substance as a function of time. In this
problem the constant C has a definite value, namely, for t = 0 we get m0  Ce 0  C .
Substituting this value of C into formula (8), we get the required dependence of the
mass of the radioactive substance on time:
m  m0 e  kt
(9)
Formula (8) is the general solution of the differential equation and formula (9)
is a particular solution which corresponds to the initial conditions of the problem.
The coefficient k can be found experimentally. For instance, for radium
k  0.000447 . The time interval T during which half the initial mass of the radioactive
substance disintegrates is called the half-life of the substance. Replacing m in
formula (9) by the value
m0
and k by the value 0.000447, we obtain an equation for
2
finding the half-life T of radium:
m0
 m0 e 0.000447T ,
2
 0.000447T   ln 2 ,
whence
35
T
ln 2
,
0.000447
T  1550 years.
II. Problem on the law of "natural growth". The law of "natural growth" is
a law according to which the rate of "growth" of a substance is directly proportional
to its amount. Let us find a formula for finding the change of the amount y of the
substance depending on time t, assuming that at the initial moment t = 0 the amount
of the substance was y0.
Here t is an independent variable and the amount of the substance at any time
moment is the required quantity. The rate of "growth" of the substance is the rate of
change of the quality y depending on the variable t .
Using the physical meaning of the derivative, as we did the preceding problem,
we can write the law of “natural growth” as follows:
dy
 ky
dt
(10)
where k  0 is the proportionality factor. Equation (10) differs from the
equation (7) only by the sign of the right-hand side and describes many processes of
“multiplication”.
The solution of equation (10), which satisfies the specified initial condition
t  0 and y  y0 , has the form
y  y 0 e kt
(11)
Formula (11) expresses the law of “natural growth”. The multiplication of
neutrons in nuclear reactions, the multiplication of bacteria, the growth of crystals –
all occur in accordance with this law.
Exercises
I. Solve these differential equations:
1.
y dx  x 2 dy  0,
11. xy /  tan y,
36
2. (1  y )dx  ( x  1)dy  0,
12. 2 xyy /  y 2  1,
3. cos 2 ydx  ( x 2  1)dy  0,
13. y /  e 2 x4 y ,
4. x 1  y 2 dx  1  x 2 dy  0,
14. y / 
5. sin u sin vdu  cos u cos vdv  0,
15. xyy/  1 ,
6. e x dx  e y (1  e x )dy  0,
16. y /  cos 2 y ,
7. 3 x  y dx  4 x  y dy  0,
17. y /  3 y 2  0 ,
8. ( xy 2  y 2 )dx  ( x 2 y  x 2 )dy  0,
18. y /  2 y  0 ,
9. ( xy 2  x)dx  ( y  x 2 y)dy  0,
19. y /  y sin x ,
10. x( y 2  1)dx  ye x dy  0,
20. 4 xdy  3 ydx  0 ,
21. x  3dy   y  1dx  0 ,
22.
1
1
dy 
dx  0 ,
cos x
sin y
23. e 2 y dx  e 3 x dy  0 ,
24.
dy
 ydx ,
x 1
2
25.
dy
 ydx  0 ,
x  2x  1
26. 2 y 2 dx  3x 3 dy  0 ,
2


27. xdy  y 2  2 y  1 dx  0 ,
29.
y
,
x3
28. cos 2 xdy  sin 2 ydx  0 ,
1
1
dx 
dy  0 ,
y
cos 5 x
30. e y 5 dx  xdy  0 .
37
Differential equations of the second order
TOPIC: Second-order equation which admit a lowering of the order. Complex
numbers. Second-order homogeneous linear differential equations with constant
coefficients
Main concepts
An equation of the form
F ( x, y, y , y )  0 ,
(1)
where x is an independent variable, y is the required function, y  and y  are
its derivatives, is called a second-order differential equation.
We study equations which can be written in the form resolved for the second
derivative:
y   f ( x, y, y ) .
(2)
The function y   (x) is called a solution of equation (1) if it’s being
substituted into the equation, turns it into an identity.
The general solution of equation (1) is a function y   ( x, C1 , C2 ) which depends
on x and on two arbitrary constants C1 and C2 if it is a solution of equation (1) for
any values of the constants C1 and C2 .
A particular solution of equation (1) is a function y   ( x, C10 , C 20 ) which results
from the general solution y   ( x, C1 , C2 ) for definite values of the constants C1  C10
and C 2  C 20 .
Second-order equation which admit a lowering of the order
An equation of the form
y   f (x )
(3)
We shall consider the particular case when by means of a change of variable
the solution of equation (1) reduces to the solution of a first-order equation. Such a
transformation of equation (1) is known as the lowering of the order.
38
The equation (3) does not include y or y  . We introduce a new function
z ( x)  y  . Then z ( x)  y  and the equation turns into a first-order equation z ( x)  f ( x)
with the required function z (x) . Solving it, we find that z ( x)   f ( x)dx  C1  F ( x)  C1 .
Since z ( x)  y  , it follows that y  F ( x)  C1 . From this, integrating once again, we get
the required solution
y   F ( x)dx  C1 x  C 2 ,
Where C1 and C2 are arbitrary constants.
Example 1. Find the general solution of the equation y   x .
Solution. Setting z ( x)  y  , we get a first-order equation z   x . Integrating it,
we find that z ( x) 
x2
 C1 . Replacing z (x ) by y  and integrating once again, we find
2
the required general solution
x2
x3
y   (  C1 )dx  C 2 
 C1 x  C 2
2
6
Example 2. Find the general solution of the equation y   cos(3x) .
Solution. Setting z ( x)  y  , we get a first-order equation z   cos(3x) . Integrating
1
3
it, we find that z ( x)  sin( 3x)  C1 . Replacing z (x) by y  and integrating once again,
we find the required general solution
1
1
y   ( sin( 3x)  C1 )dx  C 2   cos(3x)  C1 x  C 2
3
9
Complex numbers
Complex numbers are an extension of the ordinary numbers used in every day
math. They have the unique property of representing and manipulating two variables
as a single quantity.
Every complex number is the sum of two components: a real part and an
imaginary part. The real part is a real number. The imaginary part is an imaginary
number, that is, the square-root of a negative number. To keep things standardized,
39
the imaginary part is usually reduced to an ordinary number multiplied by the squareroot of negative one. As an example, the complex number
k  10   9 is first
reduced to k  10  9   1 , and then to the final form k  10  3   1 . The real part of
this complex number is 10, while the imaginary part is 3   1 . This notation allows
the abstract term,
i to denote
 1 , to be given a special symbol. Mathematicians have long used
1 .
In common case, the form of the complex number is k    i  
The mathematical notation for separating a complex number into its real and
imaginary parts uses the operators: Re and Im .
The complex numbers k1    i and k 2    i are called complex conjugate.
Just as real numbers are described as
Im
having positions along a number line,
k1  5  i
complex numbers are represented by locations
Re
in a two-dimensional display called the
complex plane.
As shown in figure, the horizontal axis
k 2  4  3i
of the complex plane is the real part of the complex number, while the vertical axis is
the imaginary part. Since real numbers are those complex numbers that have an
imaginary part equal to zero, the real number line is the same as the x-axis of the
complex plane.
Second-order homogeneous linear differential equations with constant
coefficients
An equation of the form
y   py   qy  0 ,
(4)
where p and q are real numbers is called a second-order homogeneous linear
differential equation.
For solving this equation we are replace y  by a k 2 , y  by a k and y by a ”1”
and the equation (4) turns into a quadratic equation:
40
k 2  pk  q  0
(5)
Equation (5) is called a characteristic equation of the given equation (4) and
has two roots. We designate them as k1 and k 2 .
1.
If the roots of the characteristic equation are real and distinct
( k1  k 2 ), then the general solution of equation (4) has the form
y  C1e k1x  C 2 e k2 x
2.
(6)
If the roots of the characteristic equation are real and equal
( k1  k 2 ), then the general solution of equation (4) has the form
y  C1e k1 x  C 2 x  e k1x
3.
(7)
If the roots of the characteristic equation are complex ( k1    i
and k 2    i ,   0 ), then the general solution of equation (4) has the form
y  e x (C1 cos( x)  C2 sin( x))
(8)
Example 3. Find the general solution of the equation y   y   2 y  0 .
Solution: The characteristic equation has the form k 2  k  2  0 .
Its roots k1  1 and k 2  2 being real and distinct. The general solution of the
equation has the form y  C1e1x  C2 e 2x .
Example 4. Find the general solution of the equation y   2 y   y  0 .
Solution: The characteristic equation has the form k 2  2k  1  0 .
Its roots k1  k 2  1 being real and equal. The general solution of the equation
has the form y  C1e1x  C2 x  e1x  e x (C1  C2 x) .
Example 5. Find the general solution of the equation y   4 y   13 y  0 .
Solution: The characteristic equation has the form k 2  4k  13  0 .
Its roots k1  2  3i and k 2  2  3i being complex.
The general solution of the equation has the form y  e 2 x (C1 cos 3x  C2 sin 3x) .
41
Exercises
I.
Find the general solution of these equations:
6. y  
3
1. y   2 x
2 . y   sin
x


y

2
sin
7.
3
2x
5
3. y   2 x
3
4. y   2 x 
5. y 
sin x
3
1
2
1
8.
y   e 5 x
9.
y   cos 3x  1
3


y


10.
x2
3 x
II. Find the general solution of these second-order homogeneous
linear equations:
1. y   7 y   8 y  0
6. y   10 y   25 y  0
2. y   4 y   5 y  0
7. y   17 y   66 y  0
3. y   8 y   7 y  0
8. y   3 y   22.5 y  0
4. y   16 y   63 y  0
9. y   8 y   16 y  0
5. y   18 y   81y  0
10. y   y   6 y  0
42
LABORATORY WORK №1
THE COEFFICIENT DEFINITION OF VISCOUSITY OF STOCS’
METHOD
The aim of this work: 1. To take possession of method of measurement of the
viscosity.
2. To estimate the error of measurement.
3. To study the theory of the viscosity of liquid.
The equipment: 1. Bodies of spherical form.
2. Micrometer.
3. Vertical tube with glycerin.
4. Vertical tube with oil.
5. One stop-watch.
Viscosity of liquids
In the real liquid between molecules act forces of attraction, which are the
reasons of internal friction. Internal friction, for example, causes the force of
resistance in the liquid, decreases the velocity in the condition of laminar current.
Newton determined, that force F of internal friction between two layers of liquid,
moved with different velocities, depends upon the nature of liquid and is directly
proportional to the area S of the layers in contact and to the gradient of velocity
dv / dz between them:
F 
dv
S,
dz
where  - coefficient of proportionality, called as coefficient of viscosity or
viscosity of liquids and depends upon their nature.
Force F acts tangentially on the surface of the layers of the liquid and is
directed in this manner that the accelerated layer moves more slowly and the delayed
layer moves with more speed. Gradient of velocity in this case characterizes the
43
quickness of velocity changing between the layers of the liquid direction
perpendicular to the flow of the liquid.
It is equal to: gradv 
dv v v2  v1
.


dz z
z
N  sec
Dyne  sec 
Unit of coefficient of velocity in SI:  2   Pa  sec in CGS: 
2
 .
 cm
 m 
N  sec 
2
.
 m 
This unit is called “poise”. Correlation between them is 1P=0,1 
In numerous liquids (water, low molecular organic combinations, real liquids,
melted metals and their salts) coefficient of viscosity depends only on the nature of
liquid and temperature (with increase temperature coefficient of viscosity decreases).
These liquids are called Newtonian.
For some liquids, commonly high-molecular (that is solution polymers) or
dispersion system (suspensions and emulsions), coefficient of viscosity depends also
on the flow conditions: pressure and gradient of velocity. Under their increase
viscosity of liquids decreases as a result of infringement of internal structure of the
liquid flow. Such liquids are called structurally viscous or not Newtonian liquids.
Their viscosity is characterized by the coefficient of viscosity which is related to the
conditions of the liquid flow (pressure, velocity).
Blood represented itself the suspension of forming elements in the albumen in
the albumen solution-Plasma. That is why it has to be related to the not Newtonian
liquids. Besides that under the blood flow in vessel it was observed the concentration
of forming elements in the center of flow, where viscosity relatively increases. But
the viscosity of blood doesn’t increase so much, and its coefficient of viscosity is a
constant value. Relative viscosity of blood normally is equal to 4,2-6 sP. At the
pathology conditions of patients it can decrease to 2-3 sP or increase 15-20 sP.
For measure the coefficient of viscosity of liquids with the small viscosity we
can use the method of capillary viscosimeter, based on the definition of the liquid
flow velocity in the capillary tube. Agreed with the formula of Gagen-Poiseulle
44

Pr 4
 Q 
8l


 , volume V of viscous liquid, flowing in the thin tube during an interval

of time, by other equal conditions (in detail, by same temperature) indirectly
proportional to the coefficient of viscosity  .
Scale for measurement of relative viscosity of blood consists of two grade scale
pipettes A and B with the identical capillaries a and b in the middle part. The ends of
the pipettes are connected with tee C, from which the rubber tube D with the glass
ending comes E. Pipettes B has the dividing crane G. Under the measurement it is
necessary to open the crane G and to suck in the air with the mouth through ending E
and to fill the pipette B with water at the level of mark 0. Then to close the crane to
fill the pipette A with the blood at the same level. Then to open the crane G and to
sick in the air, simultaneously sucking in the liquids from the both pipettes in this
way that blood reaches at the mark 1. The water as less viscous liquid comes to more
high mark which shows the related viscosity of blood (ill. 1)
Ill 1.
For the more viscous liquids is used the method, based on the measurement of
the bodies velocity, falling in liquids. These bodies have spherical forms (Stock’s
method). Stocks explicably observed that the body of spherical forms in the viscous
liquid is subjected to the force FC of resistance, directly proportional to velocity v ,
radius r of body and coefficient of viscosity  of liquids.
FC  6rv
45
The finding of FC was based on the way of measurement of velocity v of a
little failing ball of radius r in the cylinder, with researched liquid.
FB
FC
P
Ill 2.
When the ball is falling, three forces act on it:
4
3
4
3
Gravitation force p  mg  Vg  r 3 g . Repulsive force FB   0 gV  r 3 g 0 .
And force of resistance when the ball moves with constant velocity.
4 3
4
r g  r 3 g 0  6rv
3
3
From this coefficient of viscosity of liquids is equal to:

2 gr 2    0  2r 2 g    0 t

9v
9l
Ill 3.
46
Carrying-out the laboratory work
1. To measure diameter of a ball three times.
2. To calculate average diameter d of a ball.
3. To throw a ball in to a vessel with a liquid. To measure time t passages by a
ball distances between labels A and B (see ill 3).
4. To measure out a distance between labels l and to calculate speed of
l
t
movement v of a ball according to the formula: v  .
5. To calculate the coefficient of viscosity using the formula:

2r 2 g    0  2r 2 g    0 t
.

9v
9l
6. To make similar measurements with three balls and to find average value
coefficient of viscosity  .
7. All measured and calculated results are put in table 1.
Table 1.
№
the
d1, m d2, m d3, m d , m r, m
number of
experience
1.
2.
3.
4.
5.
6.
t, s
8. Calculated errors of measurements:




r l t

 ;
r
l
t
r  105 м ; l  5  103 мм ; t  0,25 c .
    
8. Write you conclusion.
47
v,
m/s

 N  sec 
 m2 



 N  sec 
 m2 





 N  sec 
 m2 


LABORATORY WORK №2
CALCULATION OF A SURFACE TENSION AND RESEARCH
DEPEND SURFACE TENSION OF THE TEMPERATURE
The aim of this work: 1. To take possession of method of measurement of the
surface tension.
2. To estimate the error of measurement.
3. To learn the influence of the surface active agent
(surfactant) on the surface tension of liquid.
4. To know the arrangement of the balance.
5. To study the theory of the properties of liquid surface
tension.
6. To learn dependence surface tension of temperature.
The equipment: 1. One bulk with some water.
2. One balance.
3. One ring at the hanger.
Surface tension phenomena
It is a well-known fact that some insects, for example a “pond skater” are able
to walk across a water surface; that a drop of water may remain suspended for some
time from a tap before falling, as if the water particles were held together in a bag,
that mercury gathers into small droplet when spilt; and that a dry steel needle may be
made, with care, to float on water. These observations suggest that the surface of a
liquid acts like an elastics skin covering the liquid or is in a state of tension.
Energy of Liquid Surface. Molecular Theory
That fact that a liquid surface is in a state of tension can be explained by the
intermolecular forces. In the bulk of the fluid, which begins only a few molecular
diameters downwards from the surface, a particular molecule such as A is surrounded
48
by an equal number of molecules on all sides. This can be seen by drawing a sphere
round A, fig. 1.
B
Liquid surface
F0
A
Fig. 1. Molecular forces in liquid
Consider now a molecular B in the surface of the liquid. There are very few
molecules on the vapor side above B compared with the liquid below as shown by
drawing a sphere round B. Thus if B is displaced very slightly upward, a resultant
attractive force F0 on B due to the large number of molecules below B, now has to be
overcome. It follows that if all the molecules in the surface were removed to infinity,
a definite amount of work would be needed. Consequently, molecules in the surface
have potential energy.
A molecule in the bulk of the liquid forms bonds with more neighbors than one
in the surface. Thus bonds must be broken, i.e. work must be done, to bong a
molecule into the surface. So molecules in the surface of the liquid have more
potential energy than those on the bulk. This potential energy is called the surface
energy.
Surface Area. Shape of Drop
The potential energy of any system in stable equilibrium is a minimum. Thus
under surface tension forces, the area of a liquid surface will have the least number of
molecules in it, that is, the surface area of a given volume of liquid is minimum.
Mathematically, it can be shown that the shape of a given volume of liquid with a
minimum surface area is a sphere. This is why raindrops and small droplet of
49
mercury are approximately spherical in shape. Forces acting along the tangent to the
surface of the liquid and approaching to decrease the surface area of a given volume
of liquid are called surface tension forces.
Surface tension definition
Since the surface of a liquid acts like an electric skin, the surface is in a state of
tension. A blown-up football bladder has a surface in a state of tension but the surface
tension of a bladder increases as the surface area increases, whereas the surface
tension of a liquid is independent of surface area. Any line in the bladder surface is
acted on by two equal and opposite forces, and if the bladder is cut with a knife the
rubber is drawn away from the cut by the two forces present and the bladder bursts
open.
The surface tension,  , of a fluid, sometimes called the coefficient of
surface tension, is defined at the force per unit length acting in the surface at
right angles to one side of a line drawn in the surface. The unit of  is Newton
by meter ( N / m ).
Surface tension  is a “force per unit length” or

F
l
(1)
Surface energy and work
The coefficient of surface tension in number is equal to work W which must be
done for isothermal (constant temperature) forming of a unit of the free surface of the
liquid, is proportional to the free surface area A :
W   A
(2)
Thus we have two equals definitions of the coefficient of surface tension:

F
W
and   .
l
A
50
The coefficient of surface tension depends of chemical composition of a liquid
and also its temperature (if temperature of a liquid is increasing, the coefficient of
surface tension is linearly decreasing).
For decreasing of surface tension of a liquid it is necessary to add special
impurities (soap, fat acids) in it, which take up position on the surface and decrease
the surface energy. These substances are called active substances (SAS).
Capillarity. Angle of contact. Wetting and non-wetting liquids. Rise and
fall of liquid in a capillary
When a capillary tube is immersed in water, and then placed vertically with one
end in the liquid, observation shows that water rises in the tube to a height above the
surface, the effect is due to surface tension. The narrower the tube, the greater is the
height to which the water rises. (Fig. 2a).
water
mercury
Fig 2a.
Fig 2b.
This phenomenon is known as capillarity and it occurs when blotting paper is
used to dry ink. The liquid rises up the pores of the paper when it is pressed on the
ink. When a capillarity tube is placed inside mercury, however, the liquid is
depressed below the outside level (Fig 2b). The depression increases as the diameter
of the capillary tube decreases.
51
Angle of contact
In the case of water in a glass capillary tube, if the glass is clean observation of
the meniscus shows it is hemispherical, that is, the glass surface is tangential to the
meniscus where the liquid touches it. In other cases, where liquids rise in a capillary
tube, the tangent BN to the liquid surface, where it touches the glass, may make an
acute angle  with the glass. (Fig. 3a).
The angle  is known as the angle of contact between the liquid and the glass,
and is always measured through the liquid. The angle of contact between two given
surface varies largely with their freshness and cleanliness. The angle of contact
between water and very clean glass is zero, but when the glass is not clean, the angle
of contact may be about 80 for example. When a capillarity tube is placed inside
mercury, observation shows that the surface of the liquid is depressed in the tube and
is convex upwards. (Fig. 3b). The tangent BN to the mercury at the point B, where
the liquid touches the glass thus makes an obtuse angle  , with the glass when
measured through the liquid.
B

N
θ
N
B
Angle of contact
Fig. 3a.
Fig. 3b.
Liquid wets a solid body if 0 0    90 0 (Fig. 3a). At   0 0 takes place the
phenomenon of the whole wetting. At 90 0    180 0 liquid does not wet a solid body,
(Fig. 3b), if   180 0 takes place the whole non-wetting.
52
Any doctor should take in account the coefficient of surface tension of
biological liquids as it may serve as a diagnostic factor:
 at jaundice the coefficient of surface tension of urine less than at normal
condition;
 at diabetes the coefficient of surface tension of blood changes because
of increase of lipase amount in blood.
Capillary phenomena
Capillary phenomena define conditions of condensation of vapors, boiling of
liquids, crystallization and etc. There is capillary condensation in wetted thin tubes
even at comparatively low air humidity. Because of it porous matters can delay a
sizeable amount of liquid from vapors, that makes cotton wool, underclothes and
bedlinen to be wet in damp rooms; to cause trouble drying of hygroscopic bodies,
helps not let go liquid from soil and etc. On the contrary non-wetting liquids do not
penetrate into porous bodies, so it is possible to explain impenetrability of birds
feather covered with fat for water.
Let’s examine the behavior of a air bubble which is in a capillary that contains
liquid. If the liquid pressure on the bubble is the same from different sides then both
meniscuses of the bubble will have the same radius r1  r2  of curvature, (see Fig.
4a).
F1
F2
F1
r1  r2
Fig. 4a.
F2
r1  r2
Fig. 4b.
At the liquid movement there is an excess pressure p from one of the sides
because of the shapes of meniscuses are deformed and their radiuses of curvature are
53
changed (see Fig. 4b.), the value of the excess pressure p from different sides is not
equal and the result of it is stopping of the liquid movement. Such phenomena can
happen in man’s circulatory system.
The air bubbles which got into man’s circulatory system can shut to a small
vessel and stop blood supply of any organ. This phenomenon is called embolism and
it leads to a bad functional disorders or even death. Such kind of embolism can
appear at wounding of big veins: some air bubbles that are in blood stream a bubble
and stop blood flow. Air bubbles can not be in veins at intravenous injections.
Carrying-out of the laboratory work
1. To learn the arrangement of the torsion balance and the rules of weighing.
2. To make the zero position of the balance.
3. To put the ring at the hanger on the liquid surface.
4. To start tearing off the ring and to remark the value of force at the moment
of tearing the ring of the water surface looking at the reading of the balance
(don’t tough the ring by fingers).
5. To calculate the surface tension of water using the formula:

F
,
2d
where d - the ring diameter, which you can see on the arrangement.
6. To include a heating element.
7. To write down values F for various temperatures, from 30 up to 60, through
10 degrees.
Torsion balance*. A delicate device for measuring small forces such as those
due to gravitation, magnetism, or electric charges. The force is caused to act at one
end of a small horizontal rod, which is suspended at the end of a fine vertical fibre.
The rod turns until the turning moment of the force is balanced by the tensional
reaction of the twisted fibre, the deflection being measured by an arrow and scale.
54
All measured and calculated results are put in table 1. (Take into account that
1mg=10-5N).
Table 1. Measurement of surface tension
№
the number
of
experience
1
2
3
tС
d, m
Fi, N
F,N

, N/m , N/m  m    
N/m
Fi , N
Table 2. Temperature dependence of surface tension
№
the number
of experience
1
tС
2
35
3
40
4
45
5
50
F
d, m
m, kg
Fi  F  Fi ;
SF 
, N/m
F, N
30
F1  F2  F3
;
3
 F
i
n 1
2
;

F
;
2d
t  , n - Student’s coefficient, if n  3 and   0,95 , then t  , n  4,3 ;
 F 
2
   ;
 F 
    .
8.
Draw the graf.   f t  .
9.
Write your conclusions.
55
F 
t , n
n
SF ;

LABORATORY WORK № 3
CALCULATION OF AN IMPEDANCE OF A BIOLOGICAL TISSUE
The aim of this work: 1. To take possession of method of measurement of the
impedance of a biological tissue.
2. To learn the dependence Z( ) for living, damaged
and dead tissue.
3. To define the impedance of the biological tissue.
4. To draw the graph of the dependence Z( ) for living,
damaged and dead tissue.
5. To describe your conclusions.
The equipment: 1. Audio-frequency generator.
2. Oscilloscope, the resistors (Rg), set of the equivalent
circuits.
Alternating current through LCR circuit
Alternating current is that current which continuously varies in magnitude
and periodically reversed its direction.
The frequency is generally expressed in hertz (Hz) or in cycles per second
(cps). The household supply of alternating current in Russia has a frequency of 50
cycles per second.
R, VR
L, VL
C, VC
Fig. 1.
Let an ohmic resistance R, a coil of
inductance L and a capacitor of
capacitance C is connected in series with a source of
alternating e.m.f.
(electromotive force) (Fig. 1). The currents in each element are in phase with each
other, although the voltages are not.
56
It is convenient to analyze an RLC circuit using a phasor diagram. Arrows
(acting like vectors) are drawn in an xy coordinate system to represent each voltage.
(These “vectors” are not “real”. Phasors are just a useful analytical tool.) The length
of each arrow represents the magnitude of the peak voltage across each element:
VL  I 0 X L  I 0L ;
VR  I 0 R ;
VC  I 0 X C  I 0
1
.
C
Voltage VR0 is in phase with the current and drawn along the positive x axis,
as is the current. Since VL leads the current by 900, it leads VR by 900, so is initially
drawn along the positive y axis. Voltage VC lags the current by 900, so lags VR by
900; hence, VC is drawn initially along the negative y axis. Such a diagram is shown
in Fig. 2.
VL
V0
VL-VC
I0 VR
VC
Fig. 2.
From Fig. 2 we see, using the Pythagorean theorem ( V0 is the hypotenuse of a
right triangle), that
V0  VR2  (VL  VC )2  I 02 R 2  ( I 0 X L  I 0 X C )2  I 0 R 2  ( X L  X C )2  I 0 R 2  (L 
The term
R 2  (L 
1 2
)
C
1 2
) .
C
represents the effective resistance offered by
circuit to the flow of alternating current through it. It is known as impedance of
RLC circuit and is denoted by Z.
If the alternation e.m.f. is applied to an electric circuit, the alternation current
produced is affected by the capacitance and inductance of the circuit as well as its
57
resistance. The consequent opposition is the reactance of the circuit and the total
opposition to current flow is the impedance Z. R- is a symbol of the pure ohmic
resistance. R is the ratio of the potential difference across a conductor to the current
flowing through it. The resistance R does not depend on frequency. If the current is
alternation, the resistance is the real part of the electrical impedance Z. Z=R+iX,
where X is the reactance. Reactance is the imaginary part of the complex impedance
Z.
X L  L is inductive reactance. Biological tissue has no XL.
XC 
1
1
. Here
is the capacitive reactance. Biological tissue has C,
C
C
therefore has both XC and impedance Z.
A conduction current is the motion of electrons and ions through the material
of the conductor. It is a current flowing in the substance (conductor). Z 
1
,  - it is

symbol of electrical conductivity. Electrical conductivity arises due to ions and free
electrons of biological tissue. The bad conductors have low electrical conductivity.
The good conductors have high electrical conductivity. The more frequency the less
impedance, the more conductivity of biological tissues. It is especially acts in the low
frequency range. Z 2  R2  X C 2 .
The impedance depends on the frequency.  depends on the frequency of
alternation current such dependence is called dispersion of conductivity (Fig. 3).
Z
tissue
damaged tissue
dead tissue
Lf=104Hz
Hf=108Hz
Fig. 3.
58
Log10 f
The conduction current may be the external EM fields and the living cells of
the biological tissues. Electric conductivity of living tissues depends on electric
property of blood, lymph, interstitial liquid.  of organs is about 104…106 times less
than that of its liquids (plasma, blood). Bad conductors are nervous, connective, fat
tissues. Very bad conductors concern a rough, filamentary, dry skin and especially
bones.
The range of frequencies, where the impedance varies, corresponds to change
of electrical conductivity. It is called the zone of  change. At increasing of current
frequency the impedance decreases because the membranes are collapse and capacity
disappears.
This curve is very abrupt for living tissue. The steep of curve is varies at
increasing of tissue damages. The characteristic of the tissue viability is denoted by
the letter K.
K
Zlf
Z hf
(lf – 104 Hz hf – 108 Hz)
If K  1, the viability falls. For the dead tissue K=1.
It is the important parameter for the decision of a question about suitability of
an organs for its transplantation. If it (K) corresponds to norm, the organs can be
replaced.
Let VR, VC be the potential differences across resistance R and capacitance C
respectively. In this case the voltage VC across the capacitor lags behind VR by 900,
but the voltage VR across the resistance is in phase with the current I . The total
voltage is V (the total potential differences across R and C) (Fig. 4).

VR=IR
VC=IX
V=IZ
C
Fig. 4.
59
The Pythagoras’ theorem shows, that the square of a hypotenuse is equal to the
sum of squares of cathetuses. V2=Vc2+VR2.
When C=0, Z=R. In this case a biological system has no impedance.
Figure 4 shows, that tan  
VC IX C
X

 C , for C-R series.
VR
IR
R
For biological tissue the phase angle  is the diagnostic factor. If frequency is
equal 1000 Hz, the norm for  is equal:
nerve of frog
 =640
muscle of rabbit
skin of body
 =650
 =550
The tissue has properties both resistor and condenser. What equivalent electric
circuit can replace it? What is the equivalent model of electric circuit for a biological
tissue?
It may by 1, or 2, or 3 circuits.
1) R-C in series
Fig. 5a.
When   0 , than Z   ; At low frequencies this model is far from a reality.
Z  R2 
1
 C2
2
1
V
1
tan   C  C 
VR
I R
CR
I
2) parallel circuits
When    , then Z  0
Z
1
1
  2C 2
2
R
Fig. 5b.
tan   CR
The experiments show, that at high frequencies Z experimental > Z calculation. At high
frequencies this model does not approach to real case. At high frequencies this
model is far from a reality.
3) The optimum is combination of these circuits.
R2
R1 > R2
The resistance R does not
Fig. 5c.
R1
depend on frequency.
60
Rheography is the diagnostic method based on registration of change of tissue
impedance during cardio – activity.
Simultaneously registers first derivative of a resistance, which characterizes a
rate of filling of vessels by blood. Using multichannel rheograph, it is possible to
estimate redistribution of blood between organs in various conditions.
Rheoencophalography is record of a rheogram of a brain. These methods are
widely used in medicine.
Carrying-out of the laboratory work
1. To study the electric circuit (Fig. 6).
2. To switch the generator and an oscilloscope.
3. To set on the generator frequency of 50 Hz. To receive on the screen of an
oscilloscope an ellipse.
2
4
1
3
Y
X
Fig. 6.
4. To switch the equivalent circuit 1 (  ).
5. To measure projections of an ellipse to axes of coordinates, values to write
down in the table.
6. To switch the equivalent circuit 2 (  ) and to repeat measurements, values
to write down in the table.
61
2y0
2x0
Fig. 7.
7. To switch the equivalent circuit 3 (  ) and again to repeat measurements,
values to write down in the table.
8. To change frequency sound generators according to the table and to execute
items 4-7.
9. To calculate values of an impedance for each of equivalent circuits for all
frequencies. For calculations to use the following formula:
Z
y0  Rg
x0
10. Draw the figure of Z(f). The frequency alters from 200 to 2000 Hz for each
equivalent circuit.
11. Write your conclusions.
Table
Frequenc
y, Hz
2x0, mm
2y0, mm
(  )
(  )
Z,

(  )
2x0, mm
2y0, mm
(  )
(  )
200
400
600
800
1000
1200
1400
1600
1800
2000
62
Z,

(  )
2x0, mm
2y0, mm
(  )
(  )
Z,

(  )
LABORATORY WORK № 4
STUDYING OF SPECTRAL LAWS WITH THE HELP OF THE
SPECTROSCOPE
The purpose of the work: 1. To study basic concepts and laws of the quantum
theory.
2. To understand essence of a method of the
spectral analysis.
3. To learn to make graduation of the spectral
device.
4. To find frequencies of spectra of emission and
absorption of various substances.
The equipment: 1. The spectral device UM-2.
2. Lamp.
3. Optical filters.
4. Solutions of painting substances.
Planck’s Quantum Hypothesis
Quantum theory has its origins in Planck’s hypothesis that molecular
oscillations are quantized: their energy E can only be integer (n) multiples of hf,
where f is the natural frequency of oscillation: E  nhf . This hypothesis explained
the spectrum of radiation emitted by (black) bodies at high temperature. Here h is a
constant, now called Planck’s constant, whose value was estimated by Planck by
fitting his formula for the blackbody radiation curve to experiment. The value
accepted today is h  6.626  1034 J  s .
Einstein proposed that for some experiments, light could be pictured as being
emitted and absorbed as quanta (particles), which we now call photons, each with
energy
E  hf
63
(1).
Rutherford’s “planetary” model of the Atom
Rutherford theorized that the atom must consist of a tiny but massive positively
charged nucleus, containing over 99.9 percent of the mass of the atom, surrounded by
electrons some distance away. The electrons would be moving in orbits about the
nucleus – such as the planets move around the Sun – because if they were a rest, they
would fall into the nucleus due to electrical attraction. Rutherford’s experiments
suggested that the nucleus must have a radius of about 10 -15 to 10-14 m. From kinetic
theory, and especially Einstein’s analysis of Brownian movement, the radius of atoms
was estimated to be about 10-10 m. Thus the electrons would seem to be at a distance
from the nucleus of about 10.000 to 100.000 times the radius of the nucleus itself.
Atomic spectra
As we know, heated solids (as well as liquids and dense gases) emit light with
a continuous spectrum of wavelengths. This radiation is assumed to be due to
oscillations of atoms and molecules, which are largely governed by the interaction of
each atom or molecule with its neighbors.
Rarefied gases can also be excited to emit light. The radiation from excited
gases had been observed early in the nineteenth century, and it was found that the
spectrum was not continuous, but discrete. Since excited gases emit light of only
certain wavelengths, when this light is analyzed through the slit of a spectroscope or
spectrometer, a line spectrum is seen rather than a continuous spectrum. The line
spectra emitted by a number of elements in the visible region are show here in Fig. 1.
The emission spectrum is characteristic of the material and can serve as a type of
“fingerprint” for identification of the gas.
hydrogen
Fig. 1.
helium
64
If a continuous spectrum passes through a gas, dark lines are observed in the
emerging spectrum, at wavelengths corresponding to lines normally emitted by the
gas. This is called an absorption spectrum, and it became clear that gases can
absorb light at the same frequencies at which they emit. Using film sensitive to
ultraviolet and to infrared light, it was found that gases emit and absorb discrete
frequencies in these regions as well as in the visible.
For our purposes here, the importance of the line spectra is that they are
emitted (or absorbed) by gases with low density. In such thin gases, the atoms are far
apart on the average and hence the light emitted or absorbed is assumed to be by
individual atoms rather than through interaction between atoms, as in a solid, liquid,
or dense gas. Thus the line spectra serve as a key to the structure of the atom: any
theory of atomic structure must be able to explain why atoms emit light only of
discrete wavelengths, and it should be able to predict what these wavelengths are.
Spectra can be not only linear and continuous, but also striped (molecular).
Molecular spectra correspond to transition of a molecule from one quantum condition
in another. Energy of a molecule will consist of the following parts:
E  Erotary  Eoscillatory  Eelectrons
(2),
where Erotary - energy of rotary movement of a molecule as single whole;
Eoscillatory - energy of oscillatory movement of nucleus of the atoms making a
molecule;
Eelectrons - energy of movement электронов.
As show experiments and theoretical calculations: Erotary  Eoscillator y  Eelectrons .
Hydrogen is the simplest atom – it has only one electron orbiting its nucleus. It
also has the simplest spectrum. The spectrum of most atoms shows little apparent
regularity. But the spacing between lines in the hydrogen spectrum decreases in a
regular way. In 1885, J.J. Balmer showed that the four visible lines in the hydrogen
spectrum (with measured wavelengths 656 nm, 486 nm, 434 nm, 410 nm) fit the
following formula
f 
с

 R(
1
1
 2),
2
2
n
65
n=3, 4, …
(3),
where n takes on the values 3, 4, 5, 6 for the four lines, and R, called the
Rydberg constant, has the value R  1.097  107 m1 . Later it was found that this
Balmer series of lines extended into the UV region, ending at   365nm , as show in
Fig. 2. Balmer’s formula also worked for these lines with higher integer values of n.
The lines near 365 nm became too close together to distinguish, but the limit of the
series at 365 nm correspond to n   (so 1
n2
 0 in the formula).
Later experiments on hydrogen showed that there were similar series of lines in
the UB and IR regions, and each series had a pattern just like the Balmer series, but at
different wavelengths. Each of these series was found to fit a formula with the same
form as (3) but with the 1
22
replaced by 1 2 , 1 2 , 1 2 , and so on. For example, the
1
3
4
so-called Lyman series contains lines with wavelengths from 91 nm to 122 nm (in
the UV region) and Fits the formula
с
1 1
 R( 2  2 ) , n=2, 3, …

1 n
And the wavelengths of the Paschen series (in the IR region) fit
с

 R(
1
1
 2 ) , n=4, 5, …
2
3
n
The spectral analysis is a method of definition of
quantitative and qualitative structure of substance on
his spectrum.
410 Violet
Presence in test of any element (the qualitative
434 Blue
analysis) is defined on presence in a spectrum of his
own lines. The quantitative analysis is based on that
fact, that intensity of spectral lines the above, than a lot
486 Blue green
of atoms contains in test.
The nuclear spectral analysis allows to find out
presence of an element even if his weight is equal 10-6 –
10-9 g.
In medicine the spectral analysis is used for detection of
microcells in tissues of an organism, for the
purposes of forensic medicine.
656 Red
Fig. 2.
66
Carrying-out of the laboratory work
Graduation of a scale of a spectroscope:
1. To switch on the neon lamp.
2. To study a spectrum of a neon.
3. To find the value of the number of divisions of a scale on a spectroscope’s
drum for each line of spectra.
4. Write the data in table 1.
5. Use the obtained data for making the diagram of the dependence of the
number of divisions on a drum from frequencies of spectral lines.
6. To find the value of  using the value of f and formula  
c
, where
f
c  3  108 м / с .
Table 1.
№ Position and color
Relative
Frequencies Number of Waveleng
n/n
of lines of a
brightness of
,
divisions
ths,
spectrum
spectral lines
Hz
on a drum
nm
Brightly - red
1
2 Red - orange, left of
two lines
10
4,671  1014
10
4,884  1014
3
Yellow
20
5,126  1014
4
Light green
6
5,556  1014
5
Green, right from
two identical lines
8
5,626  1014
6
Green, right from
five lines
5
5,963  1014
7
Turquoise
8
6,124  1014
8
Violet
5
6,452  1014
67
Studying of a spectrum of emission:
1. To replace a neon lamp with a torch.
2. To place in a flame of a torch some NaCl crystals.
3. To find the position of a color spectral line on a scale of a spectroscope.
4. To find the value of wavelengths of a yellow line in a spectrum of Na, using
the diagram received earlier.
5. Write down the results in the table 2.
Table 2.
Substance
Color of lines of a
spectrum
Number of
divisions on a
drum
Wavelengths,
nm
Studying of spectra of absorption:
1. To replace a torch with a usual electric lamp.
2. To place KMnO4 water solution between a lamp and a spectroscope.
3. To study a spectrum of the absorption.
4. Find the value of wavelengths of the absorbed sites of a spectrum, with help
of way described above.
5. Write down the results in the table 3.
Table 3.
Substance
Interval of divisions of a scale of the
drum, corresponding to borders of
dark sites of a spectrum.
6. Write your conclusions.
68
Interval of wavelengths,
nm
LABORATORY WORK № 5
STUDYING OF AN ELECTROSTATIC FIELD
The purpose of the work: 1. To study an arrangement of lines of force and
equipotential lines of an electric field by the
example of a dipole.
The equipment: 1. Plastic bath.
2. Electrodes.
3. Generation of a direct current.
4. Potentiometer.
5. Galvanometer.
The basic concepts and laws
There are two kinds of electric charge, positive and negative. These
designations are to be taken algebraically – that is, any charge is plus or minus so
many coulombs (C), in SI units.
Electric charge is conserved: if a certain amount of any type of charge is
produced in a process, an equals amount of the opposite type is also produced; thus
the net charge produced is zero.
According to the atomic theory, electricity originates in the atom, which
consists of a positively charged nucleus surrounded by negatively charged electrons.
Each electron has a charge - e  1.6  1019 C .
Conductors are those materials in which many electrons are relatively free to
move, whereas electric insulators are those in which very few electrons are free to
move.
An object is negatively charged when it has an excess of electrons, and
positively charged when it has less than its normal amount of electrons. The charge
on any object is thus a whole number times +e or –e; that is, charge is quantized.
An object can become charged by rubbing ( in which electrons are transferred
from one material to another), by conduction (which is transfer of charge from one
69
charged object to another by touching), or by induction (the separation of charge
within an object because of the close approach of another charged object but without
touching).
Electric charges exert a force on each other. If two charges are of opposite
types, one positive and one negative, they each exert an attractive force on the other.
If the two charges are the same type, each repels the other.
The magnitude of the force one point charge exert on another is proportional to
the product of their charges, and inversely proportional to the square of the distance
between them:
F k
Q1Q2
;
r2
this is Coulomb’s law. In SI units, k is often written as 1 4 .
0
We think of an electric field as existing in space around any charge or group
of charges. The force on another charged object is then said to be due to the electric
field present at its location.

The electric field, E , at any point in space due to one or more charges, is
defined as the force per unit charge that would act on a test charge q placed at that
point:


F
E .
q
Electric fields are represented by electric field lines that start on positive
charges and end on negative charges. Their direction indicates the direction the force
would be on a tiny positive test charge placed at a point. The lines can be drawn so

that the number per unit area is proportional to the magnitude of E ( Fig. 1a,b).
The static electric field (that is, no charges moving) inside a good conductor is
zero, and the electric field lines just outside a charged conductor are perpendicular to
its surface.
70
+
+
+
Fig. 1a.
-
Fig. 1b.
We defined the electric field as the force per unit charge. Similarly, it is useful
to define the electric potential as the potential energy per unit charge. Electric
potential is given the symbol V. If a point charge q has electric potential energy PEa
at some point a, the electric potential Va at this point is
Va 
PEa
.
q
Only differences in potential energy are physically measurable. Since the
difference in potential energy, PEa  PEb , is equal to the negative of the work, Wba,,
done by the electric force to move the charge from point b to point a, we have that the
potential difference is:
Vab  Va  Vb  
Wba
.
q
The effects of any charge distribution can be described either in terms of
electric field or in terms of electric potential. Electric potential is often easier to use
since it is a scalar, whereas electric field is a vector. There is an intimate connection
between the potential and the field. Let us examine this relation for the case of
uniform electric field, such as that between the parallel plates of Fig. 2 whose
difference of potential is Vba.
The work done be the electric field to move a positive charge q from b to a is:
W  qVba .
71
We can also write the work done as the force times distance and recall that the
force on q is F  qE , where E is the uniform electric field between the plates. Thus
W  Fd  qEd ,
where d is the distance between points a and b. We
Vba
now set these two expressions for W equal and find
qVba  qEd ,

or Vba  Ed .
Е
If we solve for E, we find that E  Vba d .
The electric potential can be represented
diagrammatically by drawing equipotential lines or,
in three dimensions, equipotential surfaces. An
equipotential surfaces is one on which al points are at
the same potential. That is, the potential difference between any
d
Fig. 2.
two points on the surface is zero, and no work is
required to move a charge from one point to the other.
An equipotential surface must be perpendicular to the electric field at any point.
The equipotential lines for the case of two
equal but oppositely charged particles are
shown in Fig. 3 as dashed lines. Equipotential
lines and surface, unlike field lines, are always
continuous and never end.
+
-
Two equals point charges q, of opposite sign,
separated by a distance l, are called an electric
dipole. The electric field lines and equipotential
surface for a dipole were shown in Fig. 3.

The product ql is called a dipole moment, p .
Fig. 3.
72
Carrying-out of the laboratory work
1. Switch on the generation of a direct current and set the voltage 1V.
2. Moving a probe in directions of coordinate axes to find 3 points for which
the current in a circuit of a probe (galvanometer) is equal to zero.
3. Put these points on a coordinate grid in a writing-book and to draw on them
a line of equals potential.
4. To set the any voltage in interval 1-5 V and to repeat items 2-3 for its.
5. Draw the electric field lines.
6. Write your conclusions.
73
LABORATORY WORK № 6
STUDYING OF A DAMPED HARMONIC MOTION WITH THE HELP OF
AN ELECTRIC RECORDER
The purpose of the work: 1. To study the basic characteristics of a damped
harmonic motion, to learn to calculate their value.
The equipment: 1. Electric recorder.
2. Spring.
3. Cargo.
4. Marker.
5. Stop watch.
6. Glue.
7. Paper.
8. Ruler.
9. The engineering micro calculator.
The basic concepts and equations
The distance x of the mass from the equilibrium point at any moment is called
the displacement. The maximum displacement – the greatest distance from the
equilibrium point – is called the amplitude, A. One cycle refers to the complete toand-fro motion from some initial point back to that same point, say from x  A to
x   A back to x  A . The period, T, is defined as the time required for one complete
cycle. Finally, the frequency, f, is the number of complete cycles per second.
Frequency is generally specified in hertz (Hz), where 1 Hz = 1 cycle per second (s-1).
Frequency and period are inversely related:
f 
1
1
and T  .
T
f
(1)
The simplest forms of oscillating are simple harmonic motion, in which the
value of physical size varies as a sinusoidal function of time.
The equation of this process is:
x  A sin( t   ) ,
(2)
where t – time,  - angular velocity (specified in radians per second). The angular
velocity can be written as
  2f 
74
2
.
T
(3)
Energy in the Simple Harmonic Oscillator
The total mechanical energy E of a mass-spring system is the sum of the
kinetic and potential energies.
The potential energy is given by
PE 
kx2
2
(4)
We can rewrite (4), using Eq. (2) as
PE 
k 2 2
A sin (t   )
2
(5)
The kinetic energy is given by
KE 
mv2
2
.
(6)
We can find the velocity v as a function of time from Eq. (2):
v
dx
 A cos(t   )
dt
(7)
We can rewrite (6), using Eq. (7) as
КЕ 
m 2 2
 A cos(t   )
2
(8)
By combining Eq. (5) with Eq. (8), we find that total mechanical energy of a
simple harmonic oscillator is equal:
W  PE  КЕ 
m 2 A 2
m
k
sin 2 t     cos 2 t      2 A 2  A 2
2
2
2


(9)
Damped harmonic motion
The amplitude of any real oscillating spring or swinging pendulum slowly
decreases in time until the oscillations stop altogether. Fig. 1 shows a typical graph of
the displacement as a function of time. This is called damped harmonic motion. The
damping is generally due to the resistance of air and to internal friction within the
oscillating system. The energy that is thus dissipated to thermal energy is reflected in
decreased amplitude of oscillation.
Fig. 1.
75
Let's consider a case, when in oscillating system are acting elastic force and
force of resistance . In this case the second law of Newton will be written down as:
m
d 2x
 kx  F ,
d t2
(10)
where F – force of resistance, k – spring coefficient
For force of resistance it is possible to write down
F  rv   r
dx
,
dt
(11)
where r – is a coefficient of resistance of environment.
We can rewrite (10), using Eq. (11) as
d 2x
dx
m
 kx  r
,
2
d t
dt
(12)
or
d 2x
dx

2

  02 x  0 ,
2
dt
d t
where 2  
(13)
r
k
and 02  (  – coefficient of damped, 0 – natural circular
m
m
frequency).
2
2
If 0    0 , then the solution of a differential equation (13) shown as:
x  Ae  t cos(t   ) ,
(14)
  t
where Ae
- amplitude of damped harmonic motion. If   0 (forces of
resistance are absent), for this case Eq. (14) will become:
x  A cos(t   ) .
The damped harmonic motion usually characterize the logarithmic decrement
of damping  :
  ln
A(t )
A(t  T )
(15)
The logarithmic decrement of damping and coefficient of damping are
connected by relation:
  T
76
(16)
Carrying-out of the laboratory work
1. To receive record of damped harmonic motion with the help of a recorder.
2. With the help of a stop watch to measure a cycle time of a drum of a
recorder.
3. With the help of a ruler to measure radius of a drum.
4. To calculate linear speed of rotation of a drum ( v 
2R
).
Trec
5. To measure length of the site L, which is corresponding to n oscillating of a
spring.
LTrec
).
n2R
1
7. To calculate frequency of damped oscillating of a spring ( f  ).
T
6. To calculate period of damped oscillating of a spring ( T 
8. To calculate circular frequency (   2f ).
9. To measure amplitude of two next damped oscillating A1 and A2.
10. To calculate decrement of damping ( d  A1 A ).
2
11. To calculate logarithmic decrement of damping (   ln d ).
12. To calculate the coefficient of damping (   T ).
Table 1.
№ Trec, V,
n/n
s m/s
n
L,
m
T,
s
f,
s-1
,
s
-1
A1,
m
A2,
m
1
2
3




Trec R L A1 A2
;




Trec
R
L
A1
A2
Trec  0.5 s ; R  L  A1  A2  0,5mm  5 10 4 m .
    
13. Write your conclusions.
77
d

,
,
-1
-1
s
s


 ,
s-1
Contents
The differential calculus (Part I)………..………………………………. 3 – 9
The differential calculus (Part II)...……..………………………………10-15
The integral calculus (Part I)…………..………………………….…..…16-21
The integral calculus (Part II)……………..…………………...…......…22-28
The differential equations of first-order…… …………………….…….29-37
The differential equations of second-order……………………….….…38-42
Laboratory work №1
The coefficient definition of viscosity of stocks’ method ……..….............43-47
Laboratory work №2
Calculation of a surface tension and research depend surface tension of the
temperature ……………………………………………………..………...48-55
Laboratory work №3
Calculation of an impedance of a biological tissue ……………..………...56-62
Laboratory work №4
Studying of spectral laws with the help of the spectroscope ……..…….…63-68
Laboratory work №5
Studying of an electrostatic field ………………………………….…..…..69-73
Laboratory work №6
Studying of a damped harmonic motion with the help of an
electric recorder……………………………………………………...…….74-78
78
Used literature
1. А.Н. Ремизов Медицинская и биологическая физика. Москва, 1996 г.
2. Н.М. Ливенцев. Курс физики для медвузов, 1978 г.
3. Практикум по МБФ под ред. проф. И.И. Маркова Ставрополь, 2001.
202 стр.
4. Shipachev V.S. Higher mathematics. M.: Mir, 1988.- 512 p.
5. Dictionary of physics. The penguin. Edited by Valery IllingWorth.
6. Douglas C. Giancoli. Physics. Principles with applications. Fifth edition.
Upper Saddle River, New Jersey 07458, 970 p.
7. Марков И.И. Курс высшей математики. Ставрополь, 2002. 88 с.
8. Dictionary of mathematics. The Penguin. Edited by John Daintish and R.D.
Nelson.
79
80