matenmat osnove fizike

8/12/2019 matenmat osnove fizike

1/23

A-1

Ta

blica SI

Conversion Factors

Length

m cm km in. ft mi

1 meter 1 102 1023 39.37 3.281 6.2143 1024

1 centimeter 1022 1 1025 0.393 7 3.2813 1022 6.2143 1026

1 kilometer 103 105 1 3.9373 104 3.2813 103 0.621 4

1 inch 2.5403 1022 2.540 2.5403 1025 1 8.3333 1022 1.5783 1025

1 foot 0.304 8 30.48 3.0483 1024 12 1 1.8943 1024

1 mile 1 609 1.6093 105 1.609 6.3363 104 5 280 1

Mass

kg g slug u

1 kilogram 1 103 6.8523 1022 6.0243 1026

1 gram 1023 1 6.8523 1025 6.0243 1023

1 slug 14.59 1.4593 104 1 8.7893 1027

1 atomic mass unit 1.6603 10227 1.6603 10224 1.1373 10228 1

Note:1 metric ton 5 1 000 kg.

Time

s min h day yr

1 second 1 1.6673 1022 2.7783 1024 1.1573 1025 3.1693 1028

1 minute 60 1 1.6673 1022 6.9943 1024 1.9013 1026

1 hour 3 600 60 1 4.1673 1022 1.1413 1024

1 day 8.6403 104 1 440 24 1 2.7383 1025

1 year 3.1563 107 5.2593 105 8.7663 103 365.2 1

Speed

m/s cm/s ft/s mi/h

1 meter per second 1 102 3.281 2.2371 centimeter per second 1022 1 3.2813 1022 2.2373 1022

1 foot per second 0.304 8 30.48 1 0.681 8

1 mile per hour 0.447 0 44.70 1.467 1

Note:1 mi/min 5 60 mi/h 5 88 ft/s.

Force

N lb

1 newton 1 0.224 8

1 pound 4.448 1

Table A.1

(Continued)

U N I V E R Z I T E T U B E O G R A D UPoljoprivredni fakultet u Zemunu

k.2015 katedra za matematiku i fiziku


2/23

A-2

Conversion Factors (continued)

Energy, Energy Transfer

J ft? lb eV

1 joule 1 0.737 6 6.2423 1018

1 foot-pound 1.356 1 8.4643 1018

1 electron volt 1.6023 10219 1.1823 10219 1

1 calorie 4.186 3.087 2.6133 1019

1 British thermal unit 1.0553 103 7.7793 102 6.5853 1021

1 kilowatt-hour 3.6003 106 2.6553 106 2.2473 1025

cal Btu kWh

1 joule 0.238 9 9.4813 1024 2.7783 1027

1 foot-pound 0.323 9 1.2853 1023 3.7663 1027

1 electron volt 3.8273 10220 1.5193 10222 4.4503 10226

1 calorie 1 3.9683 1023 1.1633 1026

1 British thermal unit 2.5203 102 1 2.9303 1024

1 kilowatt-hour 8.6013 105 3.4133 102 1

Pressure

Pa atm

1 pascal 1 9.8693 1026

1 atmosphere 1.0133 105 1

1 centimeter mercurya 1.3333 103 1.3163 1022

1 pound per square inch 6.8953 103 6.8053 1022

1 pound per square foot 47.88 4.7253 1024

cm Hg lb/in.2 lb/ft2

1 pascal 7.5013 1024 1.4503 1024 2.0893 1022

1 atmosphere 76 14.70 2.1163 103

1 centimeter mercurya 1 0.194 3 27.85

1 pound per square inch 5.171 1 144

1 pound per square foot 3.5913 1022 6.9443 1023 1

aAt 08C and at a location where the free-fal l acceleration has its standard value, 9.806 65 m/s2.

Table A.1

Symbols, Dimensions, and Units of Physical QuantitiesCommon Unit in Terms of

Quantity Symbol Unit a Dimensionsb Base SI Units

Acceleration aS

m/s2 L/T2 m/s2

Amount of substance n MOLE mol

Angle u, f radian (rad) 1

Angular acceleration aS

rad/s2 T22 s22

Angular frequency v rad/s T21 s21

Angular momentum L kg ?m2/s ML2/T kg ?m2/s

Angular velocity vS

rad/s T21 s21

Area A m2 L2 m2

Atomic number Z

Capacitance C farad (F) Q 2T2/ML2 A2?s4/kg ?m2

Charge q, Q, e coulomb (C) Q A ?s

Table A.2

(Continued)


3/23

Tables A-3

Symbols, Dimensions, and Units of Physical Quantities (continued)Common Unit in Terms of

Quantity Symbol Unit a Dimensionsb Base SI Units

Charge densityLine l C/m Q /L A ?s/m

Surface s C/m2 Q /L2 A ?s/m2

Volume r C/m3 Q /L3 A ?s/m3

Conductivity s 1/V?m Q2T/ML3 A2?s3/kg ?m3

Current I AMPERE Q /T A

Current density J A/m2 Q/TL2 A/m2

Density r kg/m3 M/L3 kg/m3

Dielectric constant k

Electric dipole moment pS

C ?m QL A ?s ?m

Electric field ES

V/m ML/QT2 kg ?m/A ?s3

Electric flux FE V?m ML3/QT2 kg ?m3/A ?s3

Electromotive force e volt (V) ML2/QT2 kg ?m2/A ?s3Energy E,U,K joule ( J) ML2/T2 kg ?m2/s2

Entropy S J/K ML2/T2K kg ?m2/s2?K

Force FS

newton (N) ML/T2 kg ?m/s2

Frequency f hertz (Hz) T21 s21

Heat Q joule ( J) ML2/T2 kg ?m2/s2

Inductance L henry (H) ML2/Q2 kg ?m2/A2?s2

Length ,,L METER L m

Displacement Dx, D rS

Distance d,h

Position x,y,z,rS

Magnetic dipole moment mS

N ?m/T QL2/T A ?m2

Magnetic field BS

tesla (T) (5Wb/m2) M/QT kg/A ?s2

Magnetic flux FB weber (Wb) ML2/QT kg ?m2/A ?s2

Mass m,M KILOGRAM M kg

Molar specific heat C J/mol ?K kg ?m2/s2?mol ?K

Moment of inertia I kg ?m2 ML2 kg ?m2

Momentum pS

kg ?m/s ML/T kg ?m/s

Period T s T s

Permeability of free space m0 N/A2 (5 H/m) ML/Q 2 kg ?m/A2?s2

Permittivity of free space P0 C2/N ?m2(5 F/m) Q 2T2/ML3 A2?s4/kg ?m3

Potential V volt (V)(5J/C) ML2/QT2 kg ?m2/A ?s3

Power P watt (W)(5J/s) ML2/T3 kg ?m2/s3

Pressure P pascal (Pa)(5 N/m2

) M/LT2

kg/m ?s2

Resistance R ohm (V)(5V/A) ML2/Q2T kg ?m2/A2?s3

Specific heat c J/kg ?K L2/T2K m2/s2?K

Speed v m/s L/T m/s

Temperature T KELVIN K K

Time t SECOND T s

Torque tS

N ?m ML2/T2 kg ?m2/s2

Velocity vS

m/s L/T m/s

Volume V m3 L3 m3

Wavelength l m L m

Work W joule ( J)(5 N ?m) ML2/T2 kg ?m2/s2

Table A.2


4/23

A-4

Matematike osnove

Ininjerske Fizike

Ovaj matematiki prirunik je deo materijala za ispit Uvod u Fiziku koji se predje na Isemestru smera za Mehanizaciju i smera za Melioracije Poljoprivrednoga fakulteta uZemunu.napomena ! .. deo ispitne literature je na engleskom jeziku...

B.11 Scientific Notation

Many quantities used by scientists often have very large or very small values. Thespeed of light, for example, is about 300 000 000 m/s, and the ink required to makethe dot over an iin this textbook has a mass of about 0.000 000 001 kg. Obviously,it is very cumbersome to read, write, and keep track of such numbers. We avoid thisproblem by using a method incorporating powers of the number 10:

1005 1

1015 10

1025 10 3 105 100

1035 10 3 10 3 105 1 000

1045 10 3 10 3 10 3 105 10 000

1055 10 3 10 3 10 3 10 3 105 100 000

and so on. The number of zeros corresponds to the power to which ten is raised,called the exponentof ten. For example, the speed of light, 300 000 000 m/s, canbe expressed as 3.003 108m/s. In this method, some representative numbers smaller than unity are thefollowing:

1021 51

105 0.1

10

22

5

1

10 3 10 5 0.01

1023 51

10 3 10 3 105 0.001

1024 51

10 3 10 3 10 3 105 0.000 1

1025 51

10 3 10 3 10 3 10 3 105 0.000 01

In these cases, the number of places the decimal point is to the left of the digit 1equals the value of the (negative) exponent. Numbers expressed as some power often multiplied by another number between one and ten are said to be in scientific

notation.For example, the scientific notation for 5 943 000 000 is 5.943 3 10

9

andthat for 0.000 083 2 is 8.32 3 1025.


5/23

B.2 Algebra A-5

When numbers expressed in scientific notation are being multiplied, the follow-ing general rule is very useful:

10n 3 10m 5 10n1m (1)

where nand m can be anynumbers (not necessarily integers). For example, 102 3105 5 107. The rule also applies if one of the exponents is negative: 103 3 1028 5 1025.

When dividing numbers expressed in scientif ic notation, note that

10n

10m5 10n 3 102m 5 10n2m .2)

Exercises

With help from the preceding rules, verify the answers to the following equations:

1. 86 400 5 8.643 104

2. 9 816 762.5 5 9.816 762 5 3 106

3. 0.000 000 039 8 5 3.98 3 1028

4. (4.03 108)(9.03 109)5 3.63 1018

5. (3.03 107)(6.03 10212)5 1.83 1024

6.75 3 10211

5.0 3 10235 1.5 3 1027

7.13 3 106 2 18 3 1022 212 3 1017 2 16 3 105 2 5 2 3 10218

B.2 Algebra

Some Basic RulesWhen algebraic operations are performed, the laws of arithmetic apply. Symbols suchas x,y, and zare usually used to represent unspecified quantities, called the unknowns.

First, consider the equation

8x5 32

If we wish to solve for x, we can divide (or multiply) each side of the equation by thesame factor without destroying the equality. In this case, if we divide both sides by 8,we have

8x

85

32

8

x5 4Next consider the equation

x 1 25 8

In this type of expression, we can add or subtract the same quantity from each side.If we subtract 2 from each side, we have

x1 2 2 2 5 8 2 2

x5 6

In general, if x 1 a5 b, then x5 b 2 a.Now consider the equation

5 5 9

1 n 1 m 1 n1m

1 n

10m1 n 1 2m 1 n2m


6/23

A-6

If we multiply each side by 5, we are left with xon the left by itself and 45 on the right:

ax515 2 5 9 3 5

x5 45In all cases, whatever operation is performed on the left side of the equality must also be per-formed on the right side.

The following rules for multiplying, dividing, adding, and subtracting fractionsshould be recalled, where a, b, c, and dare four numbers:

Rule Example

Multiplying aabb a c

db 5ac

bda2

3b a4

5b 5 8

15

Dividing1a/b21c/d2 5adbc 2/34/5 5 12 2 15 214 2 13 2 5 1012

Addinga

b6

c

d5

ad6bc

bd

2

32

4

5512 2 15 2 214 2 13 2

13

2 15

25 2

2

15

Exercises

In the following exercises, solve for x.

Answers

1. a51

1 1xx5

1 2a

a

2. x2 5 5 1 x5

3. ax2 5 5bx1 2 x57

a2b

4.5

2x1 6

53

4x1 8

x5 211

7

Powers

When powers of a given quantity xare multiplied, the following rule applies:

xnxm 5xn1m (B.3)

For example, x2x45 x2145 x6.When dividing the powers of a given quantity, the rule is

xn

xm5xn2m (B.4)

For example, x8/x25 x8225 x6.

A power that is a fraction, such as 13 , corresponds to a root as follows:

x1/n 5"n x (B.5)For example, 41/3 5"3 4 5 1.587 4 . (A scientific calculator is useful for suchcalculations.)

Finally, any quantity xnraised to the mth power is

1xn2m 5xnm (B.6)Table B.1 summarizes the rules of exponents.

Exercises

Verify the following equations:

1. 32

3 33

5 2432. x5x285 x23

nxm 5xn1m

xn

m 5 n2m

x n 5"x

1 n2m 5 nm

Table B.1 Rules of

Exponentsx0 5 1

x1 5x

xnxm 5xn1m

xn/xm 5xn2m

x1/n 5"n x1xn2m 5xnm


7/23

A-7

3. x10/x255 x15

4. 51/35 1.709 976 (Use your calculator.)

5. 601/45 2.783 158 (Use your calculator.)

6. (x4)35 x12

Factoring

Some useful formulas for factoring an equation are the following:

ax1ay1az5a1x1y1z2 common factora2 1 2ab1b2 51a1b22 perfect squarea2 2b2 51a1b2 1a2b2 differences of squares

Quadratic Equations

The general form of a quadratic equation is

ax2 1bx1c5 0 (B.7)

where xis the unknown quantity and a, b, and care numerical factors referred to ascoefficientsof the equation. This equation has two roots, given by

x52b6"b2 2 4ac

2a(B.8)

If b2$4ac, the roots are real.

2 1 x1

6"2

Example B.1

The equation x215x145 0 has the following roots corresponding to the two signs of the square-root term:

x525 6"52 214 2 11 2 14 2

2 11 2 5 25 6"92 5 25 6 32x1 5

25 1 3

25 21 x2 5

25 2 3

25 24

where x1refers to the root corresponding to the positive sign and x2refers to the root corresponding to the negative

sign.

Exercises

Solve the following quadratic equations:

Answers

1. x2 1 2x2 3 5 0 x1 5 1 x2 523

2. 2x2 2 5x1 2 5 0 x1 5 x2 512

3. 2x2 2 4x2 9 5 0 x1 5 1 1"22/2 x2 51 2"22/2Linear Equations

A linear equation has the general form

y5mx1b (B.9)y5mx1


8/23

A-8

Example B.2

Solve the two simultaneous equations

(1) 5x1y5 28

(2) 2x2 2y5 4

From Equation (2), x5y 12. Substitution of this equation into Equation (1) gives

5

1y1 2

21y5 28

6y5 218

S O L U T I O N

where m and b are constants. This equation is referred to as linear because thegraph ofyversus xis a straight line as shown in Figure B.1. The constant b, calledthe y-intercept,represents the value of yat which the straight line intersects the y

axis. The constant mis equal to theslope of the straight line. If any two points onthe straight line are specified by the coordinates (x1,y1) and (x2,y2) as in Figure B.1,the slope of the straight line can be expressed as

Slope 5y2 2y1

x2 2x15

Dy

Dx(B.10)

Note that mand bcan have either positive or negative values. If m . 0, the straightline has a positiveslope as in Figure B.1. If m , 0, the straight line has a negativeslope. In Figure B.1, both mand bare positive. Three other possible situations areshown in Figure B.2.

Exercises

1. Draw graphs of the following straight lines: (a)y 5 5x 1 3 (b)y 5 22x 1 4(c)y 5 23x 2 6

2. Find the slopes of the straight lines described in Exercise 1.

Answers (a) 5 (b) 22 (c) 23

3. Find the slopes of the straight lines that pass through the following sets ofpoints: (a) (0, 24) and (4, 2) (b) (0, 0) and (2, 25) (c) (25, 2) and (4, 22)

Answers (a) 32 (b) 252 (c) 2

49

Solving Simultaneous Linear Equations

Consider the equation 3x 1 5y5 15, which has two unknowns, xand y.Such an

equation does not have a unique solution. For example, (x5 0,y5 3), (x5 5,y5 0),and (x5 2,y5 95 ) are all solutions to this equation. If a problem has two unknowns, a unique solution is possible only if we have twopieces of information. In most common cases, those two pieces of information areequations. In general, if a problem has nunknowns, its solution requires nequations.To solve two simultaneous equations involving two unknowns, xandy, we solve one ofthe equations for xin terms ofyand substitute this expression into the other equation. In some cases, the two pieces of information may be (1) one equation and (2) acondition on the solutions. For example, suppose we have the equation m53nandthe condition that mand nmust be the smallest positive nonzero integers possible.Then, the single equation does not allow a unique solution, but the addition of thecondition gives us that n51 and m53.

lope

y

(x1,y1)

(x2,y2)

y

x(0, b)

(0, 0)x

Figure B.1 A stra ight linegraphed on an xycoordinate sys-tem. The slope of the line is theratio of Dyto Dx.

y

(1)

(2)

(3)

m 0b0

m 0b0

m 0b0

x

Figure B.2 The brown line hasa positive slope and a negativey-intercept. The blue line hasa negative slope and a positivey-intercept. The green line hasa negative slope and a negativey-intercept.


9/23

A-9

y5 23

x5y1 2 5 21

Alternative Solution Multiply each term in Equation (1) by the factor 2 and add the result to Equation (2):

10x1 2y5 216

2x2 2y5 4

12x 5 212x5 21

y5x2 2 5 23

Two linear equations containing two unknowns can also be solved by a graphi-cal method. If the straight lines corresponding to the two equations are plotted ina conventional coordinate system, the intersection of the two lines represents thesolution. For example, consider the two equations

x2y5 2

x2 2y5 21

These equations are plotted in Figure B.3. The intersection of the two lines has thecoordinates x5 5 and y5 3, which represents the solution to the equations. Youshould check this solution by the analytical technique discussed earlier.

Exercises

Solve the following pairs of simultaneous equations involving two unknowns:

Answers

1. x1y5 8 x5 5,y5 3

x2y5 2

2. 982 T5 10a T5 65, a5 3.27

T2 495 5a

3. 6x1 2y5 6 x 5 2,y5 23

8x2 4y5 28

Logarithms

Suppose a quantity xis expressed as a power of some quantity a:

x5ay (B.11)

The number a is called the basenumber. The logarithmof xwith respect to thebase a is equal to the exponent to which the base must be raised to satisfy theexpression x5 ay:

y5 logax (B.12)

Conversely, the antilogarithmofyis the number x:

x5 antilogay (B.13)

In practice, the two bases most often used are base 10, called the commonloga-rithm base, and base e5 2.718 282, called Eulers constant or the naturallogarithmbase. When common logarithms are used,

y5 log10 x 1or x5 10y2 (B.14)

y

y5 ogax

x5 antilogay

log 0 1or 1 y2

5

4

3

2

1

x2y1

2 3 4 5 6

(5, 3)

x

xy2

y

1

Figure B.3 A graphical solutionfor two linear equations.

B.2c o n t i n u e d


10/23

A-10

When natural logarithms are used,

y5 ln x 1or x5ey2 (B.15)For example, log10525 1.716, so antilog101.7165 10

1.716

5 52. Likewise, ln 52 53.951, so antiln 3.9515 e3.9515 52.In general, note you can convert between base 10 and base ewith the equality

ln x512.302 585 2 log10x (B.16)Finally, some useful properties of logarithms are the following:

log 1ab2 5 log a1 log blog 1a/b2 5 log a2 log b

log 1an2 5nlog aln e5 1

ln ea 5a

ln a1ab 5 2ln a

B.3 Geometry

Thedistance dbetween two points having coordinates (x1,y1) and (x2,y2) is

d5"1x2 2x1 22 11y2 2y1 22 (B.17)Two angles are equal if their sides are perpendicular, right side to right side and

left side to left side. For example, the two angles marked uin Figure B.4 are the samebecause of the perpendicularity of the sides of the angles. To distinguish the left andright sides of an angle, imagine standing at the angles apex and facing into the angle.

Radian measure:The arc length sof a circular arc (Fig. B.5) is proportional tothe radius rfor a fixed value of u(in radians):

s5ru

u 5s

r

(B.18)

Table B.2 gives the areasandvolumesfor several geometric shapes used through-out this text.

The equation of astraight line (Fig. B.6) is

y5mx1b (B.19)

where bis they-intercept and mis the slope of the line.The equation of acircle of radius Rcentered at the origin is

x2 1y2 5R2 (B.20)

The equation of anellipse having the origin at its center (Fig. B.7) is

x2

a21

y2

b25 1 (B.21)

where ais the length of the semimajor axis (the longer one) and bis the length ofthe semiminor axis (the shorter one).

The equation of aparabola the vertex of which is aty5 b(Fig. B.8) is

y5ax2 1b (B.22)

ln 1 y2

n 1 .30 585 2 log1log 1a2 log 1 log

log 1a 2 5 log a loglog 1 n2 og

ln 5 1

n a

n a b 5 2 na

any base

15253

B.3

5"1x x2 2 11y2 y2 2

5s

m 1

2 1 2 2

x2

a21

2

1

a 1

u

u

Figure B.4 The angles areequal because their sides areperpendicular.

r

su

Figure B.5 The angle uin radi-ans is the ratio of the arc length sto the radius rof the circle.

b

0

y

mslope

x

Figure B.6 A straight line with aslope of mand ay-intercept of b.

y

0

b

a x

Figure B.7 An ellipse w ith semi-major axis aand semiminor axis b.


11/23

B.4 Trigonometry A-11

The equation of arectangular hyperbola (Fig. B.9) is

xy5 constant (B.23)

B.4 Trigonometry

That portion of mathematics based on the special properties of the right triangleis called trigonometry. By definition, a right triangle is a triangle containing a 908angle. Consider the right triangle shown in Figure B.10, where side ais opposite the

angle u, side bis adjacent to the angle u, and side cis the hypotenuse of the triangle.The three basic trigonometric functions defined by such a triangle are the sine(sin), cosine (cos), and tangent (tan). In terms of the angle u, these functions aredefined as follows:

sin u 5side opposite u

hypotenuse5

a

c(B.24)

cos u 5side adjacent to u

hypotenuse5

b

c(B.25)

tan u 5side opposite u

side adjacent to u5

a

b(B.26)

The Pythagorean theorem provides the following relationship among the sidesof a right triangle:

c2 5a2 1b2 (B.27)

From the preceding definitions and the Pythagorean theorem, it follows that

sin2u 1 cos2u 5 1

tan u 5sin u

cos u

The cosecant, secant, and cotangent functions are defined by

csc u 5 1sin u sec u 5 1cos u cot u 5 1tan u

xy5 cons an

B.4

s n 5side opposite

hypotenuse

a

5side ad acent to

ypotenuse

an 5side opposite

side ad acent to5

c2 5a2 1 2

y

b

0x

Figure B.8 A parabola with itsvertex aty5 b.

0

y

x

Figure B.9 A hyperbola.

aopposite sidebadjacent sidechypotenuse

90uca

b

90u

Figure B.10 A right triangle,used to define the basic functionsof trigonometry.

Table B.2 Useful Information for Geometry

Surface area 2(hwhw)Volume wh

Area w

Area bh12

Area pr2

Circumference 2pr

Surface area 4pr2

Volume 4pr3

3

Volume pr2

Lateral surfacearea 2pr

Shape Area or Volume Area or VolumeShape

Sphere

r

Cylinder

Rectangular box

r

wh

Triangle

h

b

Rectangle

w

r

Circle


12/23

A-12

The following relationships are derived directly from the right triangle shown inFigure B.10:

sin u 5 cos 1908 2 u 2cos u 5 sin 1908 2 u 2cot u 5 tan 1908 2 u 2

Some properties of trigonometric functions are the following:

sin 12u 2 5 2sin ucos

12u

25 cos u

tan 12u 2 5 2tan uThe following relationships apply to anytriangle as shown in Figure B.11:

a 1 b 1 g 5 1808

a2 5b2 1c2 2 2bccos a

Law of cosines b2 5a2 1c2 2 2accos b

c2 5a2 1b2 2 2abcos g

Law of sinesa

sin a5

sin b5

c

sin g

Table B.3 lists a number of useful tr igonometric identities.

15253

ab

c

b a

g

Figure B.11 An arbitrary, non-right triangle.

Example B.3

Consider the right triangle in Figure B.12 in which a5 2.00, b5 5.00, and cis

unknown. From the Pythagorean theorem, we have

c2 5a2 1b2 5 2.002 1 5.002 5 4.00 1 25.0 5 29.0

c5"29.0 5 5.39To find the angle u, note that

tan u 5ab

5 2.005.00

5 0.400

a 2.00

b 5.00

c

u

Figure B.12 (Example B.3)

Table B.3 Some Trigonometric Identities

sin2u 1 cos2u 5 1 csc2u 5 1 1 cot2u

sec2

u 5 1 1 tan2

u sin2

u

2 51

2

11 2 cos u

2sin 2u 5 2 sin ucos u cos2u2

5 12 11 1 cos u 2cos 2u 5 cos2u 2 sin2u 1 2 cos u 5 2 sin2

u

2

tan 2u 52 tan u

1 2 tan2utan

u

25 1 2 cos u1 1 cos u

sin 1A6B2 5 sin Acos B6 cos Asin Bcos 1A6B2 5 cos Acos B7 sin Asin Bsin A6 sin B5 2 sin 3 12 1A6B2 4cos 3 12 1A7B2 4cos A1 cos B5 2 cos

32

1A1B

2 4cos

32

1A2B

2 4cos A2 cos B5 2 sin 312 1A1B2 4sin 3

12 1B2A2 4


13/23

A-13

Exercises

1. In Figure B.13, identify (a) the side opposite u(b) the side adjacent to fandthen find (c) cos u, (d) sin f, and (e) tan f.

Answers (a) 3 (b) 3 (c) 45 (d)45 (e)

43

2. In a certain right triangle, the two sides that are perpendicular to eachother are 5.00 m and 7.00 m long. What is the length of the third side?

Answer 8.60 m

3. A right triangle has a hypotenuse of length 3.0 m, and one of its angles is308. (a) What is the length of the side opposite the 308angle? (b) What isthe side adjacent to the 308angle?

Answers (a) 1.5 m (b) 2.6 m

B.5 Series Expansions

1a1b2n 5an 1 n1!

an21b1n1n2 1 2

2! an22b2 1P

11 1x2n 5 1 1nx1 n1n2 1 22!

x2 1P

ex 5 1 1x1 x2

2!1

x3

3!1P

ln 11 6x2 5 6x2 12x2 6 13x3 2Psin x5x2

x3

3!1

x5

5!2P

cos x5 1 2x2

2!1

x4

4!2P

tan x5x1 x3

31

2x5

151P 0 x0 , p

2

For x ,, 1, the following approximations can be used:1

11 1x2n < 1 1nx sin x< xex < 1 1x cos x< 1

ln 11 6x2 < 6x an x< xB.6 Differential Calculus

In various branches of science, it is sometimes necessary to use the basic tools ofcalculus, invented by Newton, to describe physical phenomena. The use of calculusis fundamental in the treatment of various problems in Newtonian mechanics, elec-tricity, and magnetism. In this section, we simply state some basic properties andrules of thumb that should be a useful review to the student.

B.5

xin radians

155626553

.6

5

4

3

u

f

Figure B.13 (Exercise 1)

1The approximations for the functions sin x, cos x, and tan xare for x#0.1 rad.

B.3c o n t i n u e dUsing a calculator, we find that

u 5tan

21

10.400

25

21.88

where tan21(0.400) is the notation for angle whose tangent is 0.400, sometimes written as arctan (0.400).


14/23

A-14

First, a function must be specified that relates one variable to another (e.g.,a coordinate as a function of time). Suppose one of the variables is called y(thedependent variable), and the other x(the independent variable). We might have a

function relationship such asy1x2 5ax3 1bx2 1cx1d

If a, b, c, and dare specified constants,ycan be calculated for any value of x.We usu-ally deal with continuous functions, that is, those for whichyvaries smoothly with x. The derivativeofywith respect to xis defined as the limit as Dxapproaches zeroof the slopes of chords drawn between two points on the y versus xcurve. Math-ematically, we write this definition as

dy

dx5 lim

DxS 0

Dy

Dx5 lim

DxS 0

y1x1 Dx2 2y1x2Dx

(B.28)

where Dyand Dxare defined as Dx5 x22 x1and Dy5y22y1(Fig. B.14). Note thatdy/dx does notmean dydivided by dx, but rather is simply a notation of the limiting

process of the derivative as defined by Equation B.28. A useful expression to remember wheny(x)5 axn, where ais a constantand nisanypositive or negative number (integer or fraction), is

dy

dx5naxn21 (B.29)

Ify(x) is a polynomial or algebraic function of x, we apply Equation B.29 to eachterm in the polynomial and take d[constant]/dx5 0. In Examples B.4 through B.7,we evaluate the derivatives of several functions.

Special Properties of the Derivative

A. Derivative of the product of two functions If a functionf(x) is given bythe product of two functionssay, g(x) and h(x)the derivative off(x) isdefined as

d

dxf 1x2 5 d

dx 3g 1x2h1x2 4 5gdh

dx1h

dg

dx(B.30)

B. Derivative of the sum of two functions If a functionf(x) is equal to thesum of two functions, the derivative of the sum is equal to the sum of thederivatives:

d

dxf 1x2 5 d

dx 3g 1x2 1h1x2 4 5 dg

dx1

dh

dx(B.31)

C. Chain rule of differential calculus Ify5f(x) and x5g(z), then dy/dzcanbe written as the product of two derivatives:

y

dz5

y

dx

dx

dz(B.32)

D. The second derivative The second derivative ofywith respect to xisdefined as the derivative of the function dy/dx(the derivative of the deriva-tive). It is usually written as

d2y

dx25

d

dx ady

dxb (B.33)

Some of the more commonly used derivatives of functions are listed in Table B.4.

y

xn2

x1 2

x3 1 2 1 2 4

x1

g

x

x1 2

x3 1 2 11 2 4 g

x1

x

y

z

y

x

x

z

2

x5

xa y

xb

Table B.4 Derivative

for Several Functionsd

dx 1a2 5 0

d

dx 1axn2 5naxn21

ddx

1 eax

2 5aeax

d

dx 1sin ax2 5acos ax

d

dx 1cos ax2 5 2asin ax

d

dx 1 tan ax2 5asec2ax

d

dx 1cot ax2 5 2acsc2ax

d

dx 1sec x2 5 tan xsec x

ddx

1csc x2 5 2cot xcsc xd

dx1 ln ax2 5 1

x

d

dx 1sin21ax2 5 a"1 2a2x2

d

dx 1cos21ax2 5 2a"1 2a2x2

d

dx 1 tan21ax2 5 a

1 1a2x2

Note: The symbols a and n represent

constants.

y

y2

y1

x1 x2x

x

y

Figure B.14 The lengths Dxand Dyare used to define the derivative ofthis function at a point.


15/23

A-15

Example B.4

Supposey(x) (that is,yas a function of x) is given by

y1x2 5ax3 1bx1cwhere aand bare constants. It follows that

y1x1 Dx2 5a1x1 Dx2 3 1b1x1 Dx2 1c5a1x3 1 3x2Dx1 3xDx2 1 Dx3 2 1b1x1 Dx2 1c

so

Dy5y1x1 Dx2 2y1x2 5a13x2Dx1 3xDx2 1 Dx3 2 1bDxSubstituting this into Equation B.28 gives

dy

dx5 limDxS 0

Dy

Dx5 limDxS 0 33ax

2

1 3axDx1aDx2

41b

dy

dx5 3ax2 1b

Example B.5

Find the derivative of

y1x2 5 8x5 1 4x3 1 2x1 7

Applying Equation B.29 to each term independently and remembering that d/dx(constant) 50, we have

dy

dx5 8 15 2x4 1 4 13 2x2 1 2 11 2x0 1 0

dy

dx5 40x4 1 12x2 1 2

S O L U T I O N

Example B.6

Find the derivative ofy(x)5 x3/(x1 1)2with respect to x.

We can rewrite this funct ion asy(x) 5x3(x11)22and apply Equation B.30:

dy

dx51x1 1 222 d

dx 1x3 2 1x3 d

dx 1x1 1 222

51x1 1 222 3x2 1x3 122 2 1x1 1 223y

dx5

3x21x1 1 22 2 2x3

1x1 1 2 3 5 x2 1x1 3 21x1 1 23

S O L U T I O N


16/23

A-16

B.7 Integral Calculus

We think of integration as the inverse of differentiation. As an example, considerthe expression

f

1x

25

y

dx5 3ax2 1b (B.34)

which was the result of differentiating the function

y1x2 5ax3 1bx1cin Example B.4. We can write Equation B.34 as dy5 f(x)dx5 (3ax21 b)dxandobtainy(x) by summing over all values of x.Mathematically, we write this inverseoperation as

y1x2 53f 1x2 dxFor the functionf(x) given by Equation B.34, we have

y

1x

253 1

3ax2 1b

2dx5ax3 1bx1c

where cis a constant of the integration. This type of integral is called an indefiniteintegralbecause its value depends on the choice of c. A generalindefinite integralI(x) is defined as

I1x2 53f 1x2 dx (B.35)wheref(x) is called the integrandandf(x) 5 dI(x)/dx. For a general continuousfunctionf(x), the integral can be interpreted geometri-cally as the area under the curve bounded byf(x) and the xaxis, between two speci-fied values of x, say, x1and x2, as in Figure B.15.

The area of the blue element in Figure B.15 is approximatelyf(xi) Dxi. If we sumall these area elements between x1and x2and take the limit of this sum as DxiS0,

B.7

1 25

y

x5 1

1x2 5 1 2 x

Example B.7

A useful formula that follows from Equation B.30 is the derivat ive of the quotient of two functions. Show that

d

dx c g 1x2

h1x2 d 5h

dg

dx2g

dh

dx

h2

We can write the quotient as gh21and then apply Equations B.29 and B.30:

d

dxa g

h5

d

dx 1gh21 2 5g d

dx 1h21 2 1h21 d

dx 1g2

5 2gh22dh

dx1h21

dg

dx

5

hdgdx

2gdhdx

h2

S O L U T I O N


17/23

A-17

we obtain the truearea under the curve bounded by f(x) and the xaxis, betweenthe limits x1and x2:

Area 5 limDxiS 0

ai

f1xi2Dxi 53x2x

1

f1x2 dx (B.36)Integrals of the type defined by Equation B.36 are called definite integrals.

One common integral that arises in practical situations has the form

3xndx5 xn11

n1 11c 1n2 21 2 (B.37)

This result is obvious, being that differentiation of the right-hand side with respectto xgivesf(x)5 xndirectly. If the limits of the integration are known, this integralbecomes a definite integraland is written

3x2

x1

xndx5 xn11

n1 1 ` x2

x15

x2n11 2x1

n11

n1 1 1n2 21 2 (B.38)

rea 5 mDxiS 0 i

1xi2D i 53x 1x2 x

n xn1

n11 1 2 1 2

x

x1

n xn

n1 1x

x

x n1 x n1

n1 11 2 1 2

xi

x2

f(xi)

f(x)

x1

Figure B.15 The definiteintegral of a function is the areaunder the curve of the functionbetween the limits x1and x2.

Examples

1.3a

0

x2dx5x

3 da

0

5a

33. 3

5

3

xdx5x2

2 d 5

3

552 2 32

25 8

2.3b

0

x3/2dx5x5/2

5/2 db

0

5 25b5/2

Partial IntegrationSometimes it is useful to apply the method of partial integration(also called inte-grating by parts) to evaluate certain integrals. This method uses the property

3 udv5uv23 vdu (B.39)where uand vare carefullychosen so as to reduce a complex integral to a simplerone. In many cases, several reductions have to be made. Consider the function

I1x2 53x2exdxwhich can be evaluated by integrating by parts twice. First, if we choose u5 x2, v5 ex,we obtain

3x2ex dx53x2d1 ex2 5x2ex 2 23exxdx1c1

3 v 3 u


18/23


19/23

A-19

Table B.6 Gausss Probability Integral and Other Definite Integrals

3`

0

xn e2ax dx5 n!

an11

I0 53`

0

e2ax2 dx5

1

2pa (Gausss probability integral)I1 53

`

0

xe2ax2

dx51

2a

I2 5

3

`

0

x2e2ax2 dx5 2

dI0

da

51

4

p

a3

I3 53`

0

x3e2ax2

dx5 2dI1

da5

1

2a2

I4 53`

0

x4e2ax2 dx5

d2I0

da25

3

8pa5I5 53

`

0

x5e2ax2

dx5d2I1

da25

1

a3

f

I2n 5121 2n dndan

I0

I2n11 5

121

2n

dn

dan I1

Table B.5 Some Indefinite Integrals (continued)

3 xdxa2 6x2

5 612ln 1a2 6x2 2 3sin2 axdx5x2 2 sin 2ax4a3 dx"a2 2x2 5 sin21

x

a5 2cos21

x

a 1a2 2x2 . 0 2 3cos2 axdx5x2 1 sin 2ax4a

3 dx"x2 6a2 5 ln 1x1"x2 6a2 2 3 dx

sin2 ax5 2

1

acot ax

3 xdx"a2 2x2 5 2"a2 2x2 3 dx

cos2 ax5

1

atan ax

3 xdx"x2 6a2 5"x2 6a2 3tan2axdx51

a 1 tan ax2 2x

3

"a2 2x2 dx5 1

2

ax

"a2 2x2 1a2sin21

x

0a

0b

3cot2axdx5 2

1

a

1cot ax

22x

3x"a2 2x2 dx5 213 1a2 2x2 2 3/2 3 sin21axdx5x1sin21ax2 1"1 2a2x2a3"x2 6a2dx5 12x"x2 6a2 6a2ln 1x1"x2 6a2 2 3 cos21axdx5x1cos21ax2 2"1 2a2x2a3x1"x2 6a2 2 dx5 13 1x2 6a2 23/2 3 dx1x2 1a2 2 3/2 5 xa2"x2 1a23eaxdx5 1aeax 3

xdx1x2 1a2 2 3/2 5 2 1"x2 1a2


20/23


21/23

A-21

Example: The Volume of a Sphere

V5 43pr3 5 43p 16.20 cm 6 2.0% 23 5 998 cm3 6 6.0%

5

1998 6 60

2cm

For complicated calculations, many uncertainties are added together, which cancause the uncertainty in the final result to be undesirably large. Experimentsshould be designed such that calculations are as simple as possible. Notice that uncertainties in a calculation always add. As a result, an experimentinvolving a subtraction should be avoided if possible, especially if the measurementsbeing subtracted are close together. The result of such a calculation is a small dif-ference in the measurements and uncertainties that add together. It is possible thatthe uncertainty in the result could be larger than the result itself!


22/23

TablicaPeriodnih

Elemenata

*Lanthanide series

**Actinide series

Atomic numberSymbol

Electron configuration

20CaAtomic mass

58

90

57

89

3

11

19

37

55

87

20

38

56

88

21

39

5771*

89103**

22

40

72

104

23

41

73

105

24

42

74

106

25

43

75

107

26

44

76

108

27

45

77

109

4

12

59 60 61 62

94939291

1

Li

Na

K

Rb

Cs

Fr

Ca

Sr

Ba

Ra

Sc

Y

Ti

Zr

Hf

Rf

V

Nb

Ta

Db

Cr

Mo

W

Sg

Mn

Tc

Re

Bh

Fe

Ru

Os

Hs

Co

Rh

Ir

Mt

Be

Mg

Ce Pr Nd Pm Sm

PuNpUPaTh

H

La

Ac

4s2

5f67s25f46d17s25f36d17s25f26d17s26d27s26d17s2

4f66s24f56s24f46s24f36s25d14f16s25d16s2

6d37s26d27s27s27s1

5d76s25d66s25d56s25d46s25d36s25d26s26s26s1

4d85s14d75s14d55s24d55s14d45s14d25s24d15s25s25s1

3d74s23d64s23d54s23d54s13d34s23d24s23d14s24s24s1

3s23s1

2s22s1

1s

(261) (262) (266) (264) (277) (268)

6.941 9.0122

1.007 9

22.990

39.098

85.468

132.91

(223)

40.078

87.62

137.33

(226)

44.956

88.906

47.867

91.224

178.49

50.942

92.906

180.95

51.996

95.94

183.84

54.938

(98)

186.21

55.845

101.07

190.23

58.933

102.91

192.2

24.305

140.12 140.91 144.24 (145) 150.36

(244)(237)238.03231.04232.04

40.078

138.91

(227)

GroupI

GroupII Transition elements

Note: Atomic mass values given are averaged over isotopes in the percentages in which they exist in nature.For an unstable element, mass number of the most stable known isotope is given in parentheses.


23/23

A-23

1.007 9

26.982 28.086 30.974 32.066 35.453 39.948

58.693

106.42

195.08

63.546

107.87

196.97

65.41

112.41

200.59

114.82

204.38

118.71

207.2

121.76

208.98

127.60

(209)

126.90

(210)

131.29

(222)

162.50 164.93 167.26 168.93 173.04

(259)(258)(257)(252)(251)

158.93

(247)

157.25

(247)

151.96

(243)

69.723 72.64 74.922 78.96 79.904 83.80

10.811 12.011 14.007 15.999 18.998 20.180

4.002 6

174.97

(262)

1

13 14 15 16 17 18

28

46

78

29

47

79

30

48

80

49

81 82 83

52

84

53

85

54

86

66 67 68 69 70

1021011009998

65

97

64

96

63

95

31 33 34 35 36

5 6 7 8 9 10

2

50 51

32

71

103

In

Ga

H

Al Si P S Cl Ar

Ni

Pd

Pt

Cu

Ag

Au

Zn

Cd

Hg Tl Pb Bi

Te

Po

I

At

Xe

Rn

Dy Ho Er Tm Yb

NoMdFmEsCf

Tb

Bk

Gd

Cm

Eu

Am

As Se Br Kr

B C N O F Ne

He

Sn Sb

Ge

Lu

Lr

GroupIII

GroupIV

GroupV

GroupVI

GroupVII

Group0

5f146d17s25f147s25f137s25f127s25f107s25f86d17s25f76d17s25f77s2

4f146s24f136s24f126s24f116s24f106s25d14f7 6s24f76s2

6p66p56p46p36p26p15d106s25d106s15d96s1

5p65p55p45p35p25p14d105s24d105s14d10

4p64p54p44p34p24p13d104s23d104s13d84s2

3p63p53p43p33p23p1

2p62p52p42p32p22p1

1s21s1

5f117s2

(271) (272) (285)

110 111 112

(289)

114

(293)

116Ds Rg Cn

(284) (288)

115113 Fl Lv

(294) (294)

5d14f8 6s2 5d14f14 6s2

Elements 113, 115, 117, and 118 have not yet been officially named. Only small numbers of atoms of these elements have been observed.Note: For a description of the atomic data, visit physics.nist.gov/PhysRefData/Elements/per_text.html.

117 118

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