lecture 1 introduction , vector calculus, functions of more variables, differential equations
DESCRIPTION
Physics for informatics. Lecture 1 Introduction , vector calculus, functions of more variables, differential equations. Ing. Jaroslav J í ra , CSc. Introduction. Lecturers: prof. Ing. Stanislav Pekárek, CSc., [email protected] , room 49A - PowerPoint PPT PresentationTRANSCRIPT
Lecture 1Introduction, vector calculus, functions of more variables,
differential equations
Ing. Jaroslav Jíra, CSc.
Physics for informatics
Introduction
Conditions for assessment:
- to gain at least 40 points,
- to measure specified number of laboratory works,
- to submit specified number of partial problems,
- to submit semester work
Source of information: http://aldebaran.feld.cvut.cz/ , section Physics for OI
Lecturers: prof. Ing. Stanislav Pekárek, CSc., [email protected] , room 49A
Ing. Jaroslav Jíra, CSc., [email protected] , room 42
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.
Textbooks: Physics I, Pekárek S., Murla M.
Physics I - seminars, Pekárek S., Murla M.
Points can be gained by:
- written tests, max. 50 points. Two tests by 25 points max. (8th and 13th week)
- semester work for max. 30 points
- activity on exercises, partial problems solving, max. 20 points
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 – first part:
Every student must solve certain number of problems according to his/her points from the semester.
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 in this way.
Then follows oral part of the exam 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 kAjAiAAAAA zyxzyx
),,(
Unit vectors i,j,k are vectors of magnitude 1 in directions of the x,y,z axes.
)1,0,0()0,1,0()0,0,1( kji
Magnitude of a vector222
zyx AAAAA
Position vector is a vector r from the origin to the current position
kzjyixzyxr
),,(
where x,y,z, are projections of r to the coordinate axes.
Adding and subtracting vectors
),,(
),,(
zyx
zyx
BBBB
AAAA
),,,( zzyyxx BABABABA
),,,( zzyyxx BABABABA
),,,( zyx AkAkAkAk
Multiplying a vector by a scalar
Example of multiplying of a vector by a scalar in a plane
)2,4()1,2(22
)1,2(
uv
u
Multiplication of a vector by a scalar in the Mathematica
Example of addition of three vectors in a plane
The vectors are given: )0,2();3,2();1,2( wvu
Numerical addition gives us
)4,2()031),2(22( wvuz
Graphical solution:
Addition of three vectors in the Mathematica
Example of subtraction of two vectors a plane
The vectors are given: )2,1();3,2( vu
Numerical subtraction gives us
)1,3()23),1(2( vuz
Graphical solution:
Subtraction of two vectors in the Mathematica
Time derivation and time integration of a vector function
ktVjtVitVVVVtV zyxzyx
)()()(),,()(
2
1
2
1
2
1
2
1
)()()()(t
t
z
t
t
y
t
t
x
t
t
dttVkdttVjdttVidttV
kdt
dVj
dt
dVi
dt
dV
dt
dV
dt
dV
dt
dV
dt
tVd zyxzyx
),,(
)(
2
1
2
1
2
1
2
1
)(,)(,)()(t
t
z
t
t
y
t
t
x
t
t
dttVdttVdttVdttV
Determine for any time t: a)
b) the tangential and the radial accelerations
Example of the time derivation of a vector
The motion of a particle is described by the vector equation
ktjtittr 32
3
1)52()(
)(),(),(),(),( tatatvtvtr
][9
1)52()( 642222 mtttzyxtr
]/[22)( 2 smktjtidt
rdtv
]/[22)( smktjdt
vdta
]/[244)( 242222 smtttvvvtv zyx
]/[1244)( 222222 smttaaata zyx
]/[2444]/[2)( 222222 smttaaasmtdt
dvta tnt
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 )(tr
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.
]/[),(),0(),()( 00 smvtgvdtgdtdtgdtgdttgv yyxyyx
]/[)81.920,10()( smttv
2
0 0 0 0( ) ( ) ( , ( ) ) ( , ) [ ]2x y y x y y
tr t v t dt v dt g t v dt v t g v t m
2( ) (10 , 20 4.905 ) [ ]r t t t t m
Time integration of a vector in the Mathematica
Projectile motion - trajectory:
Study of balistic projectile motion, when components of initial velocity are given
Scalar product (dot product) – is defined as
Where Θ is a smaller angle between vectors
a and b and S is a resulting scalar. Sbaba
Sbaba
i
n
ii
1
cos
Scalar product
cos abbabababaS zzyyxx
For three component vectors we can write
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.
abbaba
baba
abba
0Basic properties of the scalar product
Vector product
nabba
sin
Basic properties of the vector product
0
baba
abbaba
abba
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
sin abbaA
kbabajbabaibaba
bbb
aaa
kji
bac
xyyxzxxzyzzy
zyx
zyx
)()()(
Component notation
Scalar product and vector product in the Mathematica
Direction of the resulting vector of the vector productcan be determined either by the right hand rule or by the screw rule
Vector triple product
)()()( baccabcba
Scalar triple product
)()()( bacacbcba
V
ccc
bbb
aaa
cba
zyx
zyx
zyx
)(
Geometric interpretation of the scalar triple product is a volume of a paralellepiped V
Scalar field and gradient
),(),( trftrS
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).
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.
z
Sk
y
Sj
x
SiSSgrad
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: )( 22
),( yxexyxh
Extremes can be found by assuming: 0)(
hgrad
In this case : 0),()(
y
h
x
hhgrad
02 )(2)( 2222
yxyx exex
h02 )( 22
yxexyy
h
)(2)( 2222
2 yxyx exe 0y
2
121 2 xx
Answer: there are two extremes )0,2
1();0,
2
1( 21
hh
Extremes of the scalar field in the Mathematica
Vector operators
Gradient
(Nabla operator) z
Sk
y
Sj
x
SiSSgrad
Laplacian2
2
2
2
2
22
z
S
y
S
x
SSgraddivSS
Divergencez
Ak
y
Aj
x
AiAAdiv zyx
Curl
y
A
x
Ak
x
A
z
Aj
z
A
y
Ai
AAAzyx
kji
AAcurl
xyzx
yz
zyx
Differential equations
The most simple differential equation: )(' xfy
Solution of such equation is Cdxxfy )(
Example: xy 2'
dx
dyy 'where
Cxy 2
Where x2+C is general solution of the differential equation
We are looking for the function )(xy
Sometimes an additional condition is given like 3)2( y
that means the function y(x) must pass through a point ]3,2[0 x
123 2 CC
We have obtained a particular solution y(x)=x2 -1
1)( 2 xxy
First order homogenous linear differential equation with constant coefficients
0' byayThe general formula for such equation is
xey
?)( xy
To solve this equation we assume the solution in the form of exponential function.
xey xey 'If then
and the equation will change into
0baafter dividing by the eλx we obtaina
b
0 xx ebea 0)( bae x
the solution isx
a
b
Cey
Where C is a constant resulting from the initial condition
aλ+b is the characteristic equation
Example of the first order LDE – RC circuit
CR uuu
Find the time dependence of the electric current i(t) in the given circuit.
1.u Ri idt
C
Now we take the first derivative of the right equation with respect to time
011
0 iRCdt
dii
Cdt
diR
characteristic equation is
RCRC
10
1
general solution is
tRCKei1
Constant K can be calculated from initial conditions. We know that
R
uKKe
R
u
R
ui RC
0
)0(
tRCe
R
ui
1
particular solution is
Solution of the RC circuit in the Mathematica
Given values are R=1 kΩ; C=100 μF; u=10 V
Second order homogenous linear differential equation with constant coefficients
0''' cybyayThe general formula for such equation is
xey To solve this equation we assume the solution in the form of exponential function:
xey xey 'If then
and the equation will change into
02 cba after dividing by the eλx we obtain
02 xxx ecebea
0)( 2 cbae x
We obtained a quadratic characteristic equation. The roots are
xey 2'' and
a
acbb
2
42
12
?)( xy
There exist three solutions according to the discriminant D acbD 42
1) If D>0, the roots λ1, λ2 are real and different
xx eCeCy 2121
2) If D=0, the roots are real and identical λ12 =λ
xx xeCeCy 21
3) If D<0, the roots are complex conjugate λ1, λ2 where α and ω are real and imaginary parts of the root
i
i
2
1
xixxixxx eKeKeKeKy 212121
)( 21xixix eKeKey
]sin)(cos)[( 2121 xKKixKKey x
formulaEulers
xixe xi sincos
)();( 212211 KKiCKKC
]sincos[)( 21 xCxCexy x
If we substitute
we obtain
Further substitution is sometimes used cos;sin 21 ACAC
]sincoscossin[)( xAxAexy x and then
sincoscossin)sin( considering formula
we finally obtain )sin()( xAexy x
where amplitude A and phase φ are constants which can be obtained from the initial conditions and ω is angular frequency.This example leads to an oscillatory motion.
This is the solution in some cases, but …
Example of the second order LDE – a simple harmonic oscillator
Evaluate the displacement x(t) of a body of mass m on a horizontal spring with spring constant k. There are no passive resistances.
xkF If the body is displaced from its equilibrium position (x=0), it experiences a restoring force F, proportional to the displacement x:
From the second Newtons law of motion we know
xmdt
xdmmaF
2
2
0 xm
kxkxxm Characteristic
equation is02
m
k
We have two complex conjugate roots with no real part m
ki12
)sin()( tAetx tThe general solution for our symbols is
No real part of λ means α=0, and omega in our casem
k
)sin()( tAtxThe final general solution of this example is
Answer: the body performs simple harmonic motion with amplitude A and phase φ. We need two initial conditions for determination of these constants.
These conditions can be for example
)cos(2)( ttx
20cos0)0cos(
A
The particular solution is
2)0(0)0( xx
From the first condition
From the second condition
22)2
0sin( AA
)2
sin(2)( ttx
Example 2 of the second order LDE – a damped harmonic oscillator
The basic theory is the same like in case of the simple harmonic oscillator, but this time we take into account also damping.
The damping is represented by the frictional force Ff, which is proportional to the velocity v.
xcdt
dxcvcFf
The total force acting on the body is xckxFkxF f
xmmaF 0 kxxcxmxckxxm
0 xm
kx
m
cx The following substitutions
are commonly used m
c
m
k 2;
02 2 xxx Characteristic equation is 02 22
2222
12 2
442
Solution of the characteristic equation
where δ is damping constant and ω is angular frequency
There are three basic solutions according to the δ and ω.
1) δ>ω. Overdamped oscillator. The roots are real and different
222
221
tt eCeCtx 2121)(
2) δ=ω. Critical damping. The roots are real and identical.
12
tt teCeCtx 21)(
3) δ<ω. Underdamped oscillator. The roots are complex conjugate.
'
'
222
221
ii
ii
)'sin()( tAetx t
Damped harmonic oscillator in the Mathematica
Damping constant δ=1 [s-1], angular frequency ω=10 [s-1]
Damped harmonic oscillator in the Mathematica
All three basic solutions together for ω=10 s-1 Overdamped oscillator, δ=20 s-1
Critically damped oscillator, δ=10 s-1
Underdamped oscillator, δ=1 s-1