Physics for informatics
Lecture 1
Introduction, vector calculus,
functions of more variables,
Ing. Jaroslav Jíra, CSc.
Introduction
Lecturers: prof. Ing. Stanislav Pekárek, CSc., pekarek@fel.cvut.cz , room 49A
Ing. Jaroslav Jíra, CSc., jira@fel.cvut.cz , room 42
Source of information: http://aldebaran.feld.cvut.cz/ , section Physics for OI
Textbooks:
Physics I, Pekárek S., Murla M.
Physics I - seminars, Pekárek S., Murla M.
Internet tests: https://fyzika.feld.cvut.cz/auth/oitest
Scoring system of the Physics for OI
The maximum reachable amount of points from semester is 100. Points from
semester go with each student to the exam, where they create a part of the final
grade according to the exam rules.
Conditions for assessment:
- to gain at least 40 points,
- to measure six laboratory experiments and submit reports from them
Points can be gained by:
- written tests, max. 50 points. Two tests by 25 points max. (8th and 13th week)
- laboratory reports, max. 30 points. The last five lab reports are marked by up
to 6 points each
- tests on the internet, max. 20 points. There are 10 electronical tests available
on the internet, each consisting of 8 questions. Correct answering of ALL eight
questions results in 2 point gain for the student.
Examination – first part:
Every student must solve certain number of problems according to his/her
points from the semester.
Number of problems to solve
Points from the semester
1
90 and more
2
75 – 89
3
65 – 74
4
55 – 64
5
less than 55
Examination - second part:
Student answers questions in written form during the written exam. The
answers are marked and the total of 30 points can be gained this way.
Then the oral part of the exam follows and each student defends a mark
according to the table below. The column resulting in better mark is taken into
account.
written exam
semester + written exam
A
excellent
1
25
120
B
very good
1-
23
110
C
good
2
20
100
D
satisfactory
2-
18
90
E
sufficient
3
15
80
Vector calculus - basics
A vector – standard notation
for three dimensions




A  ( Ax , Ay , Az )  Ax i  Ay j  Az k
Unit vectors i,j,k are vectors of magnitude 1 in directions of the x,y,z
axes.



i  (1,0,0)
Magnitude of a vector
j  (0,1,0) k  (0,0,1)

2
2
2
A  A  Ax  Ay  Az
Position vector is a vector r from
the origin to the current position




r  ( x, y , z )  x i  y j  z k
where x,y,z, are projections of r
to the coordinate axes.
Adding and subtracting vectors
 
A  B  ( Ax  Bx , Ay  By , Az  Bz ),
 
A  B  ( Ax  Bx , Ay  By , Az  Bz ),

A  ( Ax , Ay , Az )

B  ( Bx , B y , Bz )
Multiplying a vector by a scalar

k  A  (k Ax , k Ay , k Az ),
Example of multiplying of a vector by a scalar in a plane

u  (2,1)


v  2  u  2  (2,1)  (4,2)
Multiplication of a vector by a scalar in the Mathematica
Example of addition of three vectors in a plane
The vectors are given:



u  (2,1); v  (2,3); w  (2,0)
Numerical addition gives us
   
z  u  v  w  (2  2  (2),1  3  0)  (2,4)
Graphical solution:
Addition of three vectors in the Mathematica
Example of subtraction of two vectors a plane
The vectors are given:


u  (2,3); v  (1,2)
Numerical subtraction gives us
  
z  u  v  (2  (1),3  2)  (3,1)
Graphical solution:
Subtraction of two vectors in the Mathematica
Time derivation and time integration of a vector function




V (t )  (Vx ,Vy ,Vz )  Vx (t )i  Vy (t ) j  Vz (t )k

dVx dVy dVz
dVx  dVy  dVz 
dV (t )
(
,
,
)
i
j
k
dt
dt dt dt
dt
dt
dt
 t2

 t2
 t2
 V (t )dt  i  Vx (t )dt  j  Vy (t )dt  k  Vz (t )dt
t2
t1
t1
t1
t1
t2
t2
 t2


 V (t )dt , V (t )dt , V (t )dt 
V
(
t
)
dt

t
t y
t z 
 t x
1
1
1
1

t2
Example of the time derivation of a vector
The motion of a particle is
described by the vector equation
 2 1 3

r (t )  (2t  5)i  t j  t k
3


r (t ), v (t ), v(t ), a (t ), a(t )
Determine for any time t: a)
b) the tangential and the radial accelerations
1
r (t )  x 2  y 2  z 2  (2t  5) 2  t 4  t 6 [m]
9 




 2

dv

dr
 2 j  2t k [m / s ]
v (t ) 
 2i  2t j  t k [m / s ] a (t ) 
dt
dt
v(t )  vx  v y  v z  4  4t 2  t 4  2  t 2 [m / s ]
2
2
2
a(t )  a x  a y  a z  4  4t 2  2 1  t 2 [m / s 2 ]
2
2
2
dv
at (t ) 
 2t [m / s 2 ]
dt
an  a 2  at  4  4t 2  4t 2  2 [m / s 2 ]
2
Time derivation of a vector in the Mathematica
Time derivation of a vector in the Mathematica -continued
What would happen without
Assuming and Refine
What would happen without
Simplify
Graphical output of the

r (t )
Example of the time integration of a vector
Evaluate the time dependence of the velocity and the position vector for the
projectile motion. Initial velocity v0=(10,20) m/s and g=(0,-9.81) m/s2.


v   g (t )dt  (  g x dt ,  g y dt )  (  0  dt ,  g y  dt )  (v0 x , g y t  v0 y ) [m / s ]

v (t )  (10,  9.81t  20) [m / s ]
t2
r (t )   v (t )dt  (  v0 x dt ,  ( g y t  v0 y )dt )  (v0 xt , g y  v0 y t ) [m]
2

r (t )  (10t ,4.905 t 2  20t ) [m / s]
Time integration of a vector in the Mathematica
Study of balistic projectile motion, when components of initial velocity are given
Projectile motion - trajectory:
Scalar product
Scalar product (dot product) – is defined as
Where Θ is a smaller angle between vectors
a and b and S is a resulting scalar.
   
a  b  a  b  cos   S
  n
a  b   ai bi  S
i 1
For three component vectors we can write
 
S  a  b  ax  bx  a y  by  az  bz  ab  cos 
Geometric interpretation – scalar product is
equal to the area of rectangle having a and
b.cosΘ as sides. Blue and red arrows
represent original vectors a and b.
Basic properties of the scalar product
   
a b  b  a
 
 
a  b  a b  0
   
a b a  b  ab
Vector product
Vector product (cross product) – is defined as
Where Θ is the smaller angle between vectors
a and b and n is unit vector perpendicular to the
plane containing a and b.
Geometric interpretation - the magnitude of the
cross product can be interpreted as the positive
area A of the parallelogram having a and b as
sides
 

a  b  ab  sin   n
 
A  a  b  ab  sin 
Component notation

i

j

k
  
c  a  b  ax
ay
az 
bx
by
bz



 (a y bz  a z by )i  (a z bx  a x bz ) j  (a x by  a y bx )k
Basic properties of
the vector product
 
 
a  b  b  a
 
 
a  b  a  b  ab
   
a b a b  0
Scalar product and vector product in the Mathematica
Direction of the resulting vector of the vector product
can be determined either by the right hand rule or by the screw rule
Vector triple product
     
  
a  (b  c )  b (a  c )  c (a  b )
Scalar triple product
     
  
a  (b  c )  b  (c  a )  c  (a  b )
ax
  
a  (b  c )  bx
cx
ay
by
cy
az
bz  V
cz
Geometric interpretation of
the scalar triple product is a
volume of a paralellepiped V
Scalar field and gradient
Scalar field associates a scalar quantity to every point in a space. This
association can be described by a scalar function f and can be also time
dependent. (for instance temperature, density or pressure distribution).


S (r , t )  f (r , t )
The gradient of a scalar field is a vector field that points in the direction of the
greatest rate of increase of the scalar field, and whose magnitude is that rate
of increase.
 S  S  S
grad S  S  i
j
k
x
y
z
Example: the gradient of the function
f(x,y) = −(cos2x + cos2y)2 depicted as
a projected vector field on the bottom
plane.
Example 2 – finding extremes of the scalar field
Find extremes of the function:
Extremes can be found by assuming:
h( x, y)  xe

grad (h)  0
( x 2  y 2 )
In this case :
h h 
grad (h)  ( , )  0
x y
h
( x 2  y 2 )
2 ( x 2  y 2 )
e
 2x e
0
x
h
( x 2  y 2 )
 2 xye
0
y
e
( x2  y 2 )
2 ( x2  y 2 )
 2x e
y0
1
1  2x  x  
2
2
Answer: there are two extremes
1
1
h1  (
,0); h2  (
,0)
2
 2
Extremes of the scalar field in the Mathematica
Vector operators
Gradient
(Nabla operator)
Divergence
Curl
 S  S  S
grad S  S  i
j
k
x
y
z

 Ax Ay Az
div A    A 


x
y
z



i
j
k

 

   Az Ay 
 
curl A    A 
 i 

x y z
z 
 y
Ax Ay Az
  Ax Az    Ay Ax 

 j


  k 
x   x
y 
 z
Laplacian
2S 2S 2S
S   S  div grad S  2  2  2
x
y
z
2
Basic mechanical quantities and relations
and their analogies in linear and rotational motion
Linear motion
s, r
path, position vector
Rotational motion
[m]
φ
angle
[ rad ]
v
velocity
[ m*s-1 ]
ω
anglular velocity
[ rad*s-1]
a
acceleration
[ m*s-2 ]
ε
angular acceleration
[ rad*s-2 ]
F
force
[N]
M
torque
m
mass
[ kg ]
J
moment of inertia
p
linear momentum
[ kg*m*s-1]
b
angular momentum
Work
W= F s
Work
W= M φ
Kinetic energy Ek= ½ m v2
Kinetic energy Ek= ½ J ω2
Equation of motion F = m a
Equation of motion M = J ε
[ N*m]
[ kg*m2 ]
[kg*m2*s-1]