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Unformatted text preview: POWERS OF TEN . You should be familiar with the usage of powers of ten. it is a compact form of
Writing very large or very small numbers. For example, instead of 10 000, we write
104, where the exponent represents the number of zeros: that is, 10“ = 10 X
10 X 10 x 10 = 10 000. Likewise, a small number like 0.0001 can be expressed as
10“, where the negative exponent indicates that we are dealing with a number less than one. Some other examples of the use of powers of ten are 1000 = 10% 0.003 = 3 x 103
85 000 = 8.5 x 104 0.00085 = 8.5 x 104
3 200000 =_ 3.2 x 106 0.00002 = 2 x 105 If numbers written as powers of ten are multiplied, we simply add the exponents,
maintaining their signs. For example, (3 X 103) x (5 x 104) = 15 x 107 = 1.5 x 105
(2 x 105) X (4 X 10‘?) = 8 X 103
(5.6 x 104) x (4.3 x 108) = 24 x 1012 When numbers written as powers of ten are divided, we can bring the power of
ten from the denominator to the numerator by changing its sign. For example, géig: = 4x105 x 102 = 4 x103
' ff—llg:=3x104x109=3x105
In general,
10m)" : 10m
10"/10"' = ionm
(10")"' = 10m
ALGEBRA When algebraic operations are performed, the laws of arithmetic apply. SymbolS
such as x, y, and z are usually used to represent quantities that are not speciﬁed, and
symbols such as a, b, and c are used to represent numbers. You should be familiar with the following operations: a(%)=%f’— (one (cl/b) _Ec_i c _ adrcb
(c/d) _. be _ Fractions 0
_+..._
b_d bd
ax+bx=x(a+b} 12—y2=(r+y)(x—y)
x2—2x——15=(x+3)(x——5) Factoring and
combinations quadratic equation _b t \/b2 — 4ac th =
En x 20 Multiplying powers Rootsofa '[Ifax'2+bx+,c=0,
of a given quantity [ ).._, ln = logarithm to base 8
log = logarithm to base 10 in e :1
lneI = x “'
ln(xy) lnx + In y Logarithmic functions ln{x/y) ___ In x _. In 9. 111(1/1') = —inx In x" = n lnx lo a = 2.302610ga
log a = 0.434291n a Simultaneous Linear Equations In order to solve two simultaneous equations involving two unknowns, x and y, we solve one of the equations for x in terms of y and substitute this expression into
the other equation. Example (1) 5x + y = 8 (2) 2x —— 2y = 4 Alternate Solution: Multiply (1) by 2 and add the result
to (2): Solution. From (2), x = y + 2. Substitution of this into (1)
gives 10:: + 2y = —16 5(y+2}+y=—8 ___L__232 =4 6y=—18 12x=—12 y:—3 x=—1
x=y+2=m1 y:x2=—3 GEOMETRY
The Pythagorean theorem, which relates the three sides of a right triangle
I The radian measure: the arc lengths is proportional to the radius r for a ﬁxed Value of 9 (in radians)
3 = E I A
* r .
Circumference of a circle .
I C = 2771' Area of a circle A = nrz
Area of a triangle
1
A = —bh ‘
2 ‘“'
Surface area of a sphere
Volume of a sphere
Volume of a cylinder
V = 1"le Equation of a straight line ! = b = y intercept
.‘y mx+b m=slope=tanﬂ Equation of a circle of radius R centered at the origin Equation of a parabola whose vertex is at y = b .— TRIGONOMETRY The sine, cosine, and tangent functions in trigonometry are deﬁned in terms of the
ratios of the sides of a right triangle: . sinB : Side opposxte 0 __ hypotenuse c0919 = ———— =
hypotenuse a
.3
side adjacent 9 b
?
a side opposite 0 t = ....
mg side adjacent 6 b From‘the above deﬁnitions and the Pythagorean theorem, it follows that
sinzﬂ + cos20 = 1 _' sinH
tanﬁ _ cost) The cosecant, secant, and cotangent functions are deﬁned by csc6=L secG: cotﬁ: 1 $1116 , cosﬁ tanﬁ The relations at the right follow directiy from the right triangle above: sinﬁ z (205(90“ — 0) cost? = sin(90° — 9) c015: tan(90°  9)
Some properties of trigonometric functions: sin(—8) = sin0 cos(—6) = 0059 tan(—6) = —tan6
Relations that apply to any triangle: a+B+y=lSO° (:2 = b2 + c2 — 21):: case:
Law of cosines 172 = (12 + 02 — 20c cosB
6‘2 = 02 + b2  2ab cosy Law of Sines {—,— = —. = .— Trimuic Honda“ ' sin’ﬁ + c0520 = 1 csczﬂ = 1 + cotzo
secza = l + tanZB sinzg = %(1 —. 5059)
511129 = 25mg cost? coszg— = %‘1 + 3:059)
c0529 = c0530 — 511126 1 — c056 = Rina; _ 2tan0 i _ 1 .. cost?
“”29 ‘ 1 — tanza “m2 ‘ 1 + c059 sin(A i B) = sinA cosB I cosA sinB
cos(A : B] = cosA cosB : sinA sinB sinA = sagas = 25in[%(A : B}]cos[—%{A ; 3)]
cosA + 0358 = 2cos[%{A + Bi]cos[%{A  3)] cosA — 5055 = 23in[%{A + B)]sin[%(B — m] {a+b)=5~+la~—lb+ (1 +x)"=1+nx+ Suic Expansion 1! x2 x3 mm —1) n{n— 1)
2! xa+m 2! e‘=1+x+—2T+§+~ 49., w'2b2+~ 9 in radians ln{1:x}=:x—é—x¢:%x3_
sinx:x—£.+£l—
cosx=1——2x;+f—!
tanx¢x+£+£€+n [xl<fr/2 " 3 15
Forx< 1,the following approadmaﬁonscanbeused:
(l+xl"::1+nx sinxzx
e':l+x (:05le lnﬁl : x) 2: ix tanxzx DIFFERENTlAL CALCULUS In various branches of science, it is sometimes necessary to use the basic tools of
calculus, first invented by Newton, to describe physical phenomena. The use of
calculus is fundamental in the treatment of various problems in newtom‘an mechan
ics, electricity, and magnetism. In this section, we simply state some basic P'DPC"
ties and “rules of thumb” that should be a useful review to the student. First, a function must be speciﬁed which relates one variable to another (Mb ‘5
coordinate as a function of time). Suppose one of the variables is called 9 like
dependent variable), the other I (the independent variable). We might have a m
tion relation such as y(x)=ax3+bx2+cy+d _
If a, b, c, and d are speciﬁed constants, then y can be calculated for any ”31“” ”f z. Figure 1 We usually deal with continuous functions, that is, those for which y varies
“smoothly" with x. . The derivative of y with respect to x is defined as the limit of the slopes of chords
drawn between two points on the y versus at curve as An: approaches zero. Mathe
matically, we write this deﬁnition as 9y— = lim 31 = lim My“ + "W ' yiﬂ dx airo Ax its'0 Ax Where Ag and Ax are deﬁned as Ax r: x2 — x1 and Jig = 92 — 9'1 (see Fig. 1).
A useful expression to remember when y(x) :: ax", where a is a constant and n is
any positive or negative number (integer or fraction), is If ytx) is a polynomial or algebraic function of x, we apply Eq. 3.2 to each term in
the polynomial and take da/dx = 0, It is important to note that dy/dx does not mean dy divided bv dx, but is simply a notation of the limiting process of the
derivative   Properties of the Derivative Derivative of the Product of Two Functions. If a function y is given by the
product of two functions, say. g(x) and h(x), then th ‘  .
e derivative of y iS deﬁned as d , d dh d
_ = _ _ g
dr ”i dx [glx’hml ‘ g3: 4*" ”a B. Derivative of the Sum of Two Functions. If a function y is equal to the
in two functions, then the derivative of the sum is equal to the sum of the deriv ti“ of
a WES:
d ,. d . , dg dh 2..., C. Chain Rule of Differential Calculus. If y :2 ﬁx) and .r is a functio
other variable 3, then dy/dx can be written as the product of two d
dy _ dy dz
dx _ dzE n of some
erivatjvg: D. The Second Derivative. The second derivative of y with respect to x is
as the derivative of the function dy/dx (or, the derivative of the derivativ
usually written deﬁned
e}. It is day (I (dy) dx2 =71: E Derivatives for Several Function  ' . d « ﬂ Eta] = 0 Emmott) _ aseczax
d (we) — nael £(cotax) = —ocsc~‘czx
E ‘ dx
_d_(en.r) _ Gen: _d_(secx} = tanxsecx
dx — dx
d l ' t — cosax «(Hi—(cscx) = —cotxcsoc
a smax — Cl dX' 1
Edicosax] = «asinax illncvt) = 3: W
Note: The letters a and n are constants. INTEGRAL CALCULUS We can think of integration as the inverse of differentiation. As an example,
ponsider the expression ﬂx) :%=Sax2 + b the result of differentiating the function
y(x)=ax3+bx+c_ We can write the ﬁrst expression dy = f(x)dx = (302:2 + b)dx and
obtain y(x) by “summing" over all values of x. Mathematically, we write this inverse operation W) = fﬂxidx
For the function f(x) given above,
y(x) = f(3ax2 + b)dx = £113 + bx + c where c is a constant of the integration. This type of integral is called an indeﬁnite
integral since its value depends on the choice of the constant c.
A general indeﬁnite integral I (x) is deﬁned as ' HI) = from d1(x} ‘where ﬁx) is called the integrand and f(x) = dx  2..., For a general continuous function f (x), the integral can be described as the area r I Under the curve bounded by f (x) and the x axis, between two speciﬁed values of' x,
say. 3:} and x1“ as in Fig. 2. for) —*l l“—
AX]:
Figure 2 The area of the shaded element is approximater fiAxi. If we sum all these area
elernents from x1 to x2 and take the limit of this sum as .311. ——> 0, we obtain the true
“8a under the curve bounded by f (x) and 1:, between the limits I] and x2: Area = lim ZfimAxi = fzzﬂxkix 1nttbgrals of the type deﬁned by Eq. B8 are called deﬁnite integrab. One of the common types of integrals that arise in practical situations has the
form f d xn+1 + { ¢ 1)
x x = ' c n — n + l '
This result is obvious since diﬁerentiation of the right~hand side with respect to it
gives Jﬁr) = x" directly. If the limits of the integration are known, this integral
becomes a deﬁnite integral and is written n+1 __ I n+1 32!! _.‘C2 '1 .
Idx—T (13¢ l} 4'1 2..., Partial Integration Sometimes it is useful to apply the method of partial integration to evaluate ' certain integrals. The method uses the property that
fade = on — udu where u and u are carefully chosen so as to reduce a complex integral to a simpfer
one. In many cases, several reductions have to be made. Consider the example I(x) = fxze‘dx This can be evaluated by integrating by parts twice. First, if we choose a = :2.
u = e‘, we get fxze’dx = x2d(e‘) = 1:28" — 2 e’xdx + 61
Now, in the second term, choose u = x, v = 6", which gives
ftze‘dx 2 x283 — 2er + Zfe‘dx + Cl
or
fxze‘dx = xze‘ — 2m" + 2e‘ + c2
The Perfect Differential Another useful method to remember is the use of the perfect diferenhﬂil P71;
we should sometimes 100k for a change of variable such that the differentia o function is the differential of the independent variable appearing m the mtetr
For example, consider the integral
ﬁx) = fcos2x sinxdx
This becomes easy to evaluate if we rewrite the differential as d(cosx} = —sinxdx.
The integral then becomes
fcos2x sinxdx = — c0522: d(cosx}
If we now change Variables, letting y = cosx, we get
3 3
fcos2xsinxdx= —fy2dy= wy:’,—+e= —CO::I +5 in, Some Indeﬁnite lntcgrals“ (an arbitrary constant should be addzd to each of theie integrals)
WWW fr'dx: nx:=1 (provided n ¢ —1) fxé’dx:%(ax—1)
dx 4 _ dx — X i :
T‘fx (ix—h“ fa+beﬂ_3_acln(a+bec)
dx 1 . 1
In + bx = 3111(0 + bx) fsmaxdx = “Ecosax
dx 1 ‘ 1 .
fm = ‘m fcosaxdx _ Esmax
a? (1:28 = %tan"% ftanaxdx = —:1]—!n(cosax) = éinhecex)
fa? (1;xe 7' glam: :: (02 _ x2 > D) fcotaxdx = %}n(sinax) 7
dx 1 x— a 1 _ 1 ax 1T
f—ﬁxz __ .. Zinx + a (x2 — 02 >0) fsecaxdx .—_ Eli:(secax + tanax) _ Eln[mn(7 + 1)]
xdx _ 1 _ 1 ax
2.2—2? = —§1“{°2 I X2) . fcscaxdx _ Elnkscax «— cotax) _ Elr'1(1:an«~2n.)
f . dx = sin1: = —cos'1£ (92 __ x2 >0} fsinzaxdx = i _ sinZax
V02 — x2 a a 2 4a
dx r—m 2 _ 5 sin2ax
fxm=lnix+yxziazj fcosaxdx_2+ 4a
dx 1
I x—Jfd—x—T = —V‘02 — x2 fsinzax :2 —Ecotax
V0? _— x“ dx 1
f ' mix = x2 i ”2 fcos2ax = Etanax
Vxe : a2 2 1
~—— tan (3de = —[tanax)  x
fv‘d’ — dex 2 %(x\/a2 — x2 + (ism1%) f a
1
fxx/a? _ xzdx = ,ﬂaz _ xszz _ fcotzaxdx = E(comx) _ x
r'—— — " '“1dxx{‘‘ ) \f1—02x2
fwa : 0de = gwxz : a2 : a21n(x + Wu f5!“ ‘1’“ — 5‘“ a" + —a
fw—‘xz
I'M/F 2 02d): =1HX2 : 02%” fcos‘laxdx = x(cos‘1ax) — LL75;
_ 1 . ,
fend): = .159: ftan"axdx = x(tan lax) — Emu + ax)
flnaxdx = (ﬁnd!) — x fem2am = x(cot"ax) + 21_ain(1 + azx'cy W ' Gauss‘ probability integral and other integrals can be found on the right iron: endpaper‘ G a
n!
I D ﬁts—MdzEG—‘tﬁ, n>—l,a>0
“ .I. It
a. rﬂ’dz=— —
o 2 6
fl! “GE, 1 r
4 3'8 dz=~ 
o 4 aa
°° 1 3 (2n 1)
5_ Bun—oz! I: ' ‘— '
j; 5" 6 dz 2n+1 ‘IGZMI
°° l
6. zewd$=—
0 20
”° 1
8“? _.._.
ti: :6 (ix—2a: +1 1
14.1.1 m—"dz=a—,{e“—e"“a(e°+e"“)} +1
15. 1 3‘9de = (—1)"+‘A.(—a) — Ada) ...
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 Summer '10
 ERBILL
 Physics

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