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., pekarek@fel.cvut.cz , room 49A

Ing. Jaroslav Jíra, CSc., jira@fel.cvut.cz , 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