Tables of math, physics and chemistry engineering handbook
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N C E E SFUNDAMENTALS OF ENGINEERINGSUPPLIED-REFERENCE HANDBOOKFIFTH EDITION N A T I O N A L C O U N C I L O F E X A M I N E R S F O R E N G I N E E R I N G A N D S U R V E Y I N G 2001 by the National Council of Examiners for Engineering and Surveying® .
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FUNDAMENTALS OF ENGINEERINGSUPPLIED-REFERENCE HANDBOOKFIFTH EDITION Prepared by National Council of Examiners for Engineering and Surveying® (NCEES®) 280 Seneca Creek Road P.O. Box 1686 Clemson, SC 29633-1686 Telephone: (800) 250-3196 Fax: (864) 654-6033 www.ncees.org
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2001 by the National Council of Examiners for Engineering and Surveying® .All rights reserved. First edition 1996Fifth edition 2001
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FOREWORDDuring its August 1991 Annual Business Meeting, the National Council of Examiners for Engineering and Surveying (NCEES)voted to make the Fundamentals of Engineering (FE) examination an NCEES supplied-reference examination. Then during itsAugust 1994 Annual Business Meeting, the NCEES voted to make the FE examination a discipline-specific examination. As aresult of the 1994 vote, the FE examination was developed to test the lower-division subjects of a typical bachelor engineeringdegree program during the morning portion of the examination, and to test the upper-division subjects of a typical bachelorengineering degree program during the afternoon. The lower-division subjects refer to the first 90 semester credit hours (fivesemesters at 18 credit hours per semester) of engineering coursework. The upper-division subjects refer to the remainder of theengineering coursework.Since engineers rely heavily on reference materials, the FE Supplied-Reference Handbook will be made available prior to theexamination. The examinee may use this handbook while preparing for the examination. The handbook contains only referenceformulas and tables; no example questions are included. Many commercially available books contain worked examples andsample questions. An examinee can also perform a self-test using one of the NCEES FE Sample Questions and Solutions books(a partial examination), which may be purchased by calling (800) 250-3196.The examinee is not allowed to bring reference material into the examination room. Another copy of the FE Supplied-ReferenceHandbook will be made available to each examinee in the room. When the examinee departs the examination room, the FESupplied-Reference Handbook supplied in the room shall be returned to the examination proctors.The FE Supplied-Reference Handbook has been prepared to support the FE examination process. The FE Supplied-ReferenceHandbook is not designed to assist in all parts of the FE examination. For example, some of the basic theories, conversions,formulas, and definitions that examinees are expected to know have not been included. The FE Supplied-Reference Handbookmay not include some special material required for the solution of a particular question. In such a situation, the required specialinformation will be included in the question statement. DISCLAIMER: The NCEES in no event shall be liable for not providing reference material to support all the questions in the FE examination. In the interest of constant improvement, the NCEES reserves the right to revise and update the FE Supplied-Reference Handbook as it deems appropriate without informing interested parties. Each NCEES FE examination will be administered using the latest version of the FE Supplied- Reference Handbook. So that this handbook can be reused, PLEASE, at the examination site, DO NOT WRITE IN THIS HANDBOOK. iii
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UNITSThis handbook uses the metric system of units. Ultimately, the FE examination will be entirely metric. However, currently someof the problems use both metric and U.S. Customary System (USCS). In the USCS system of units, both force and mass arecalled pounds. Therefore, one must distinguish the pound-force (lbf) from the pound-mass (lbm).The pound-force is that force which accelerates one pound-mass at 32.174 ft/s2. Thus, 1 lbf = 32.174 lbm-ft/s2. The expression32.174 lbm-ft/(lbf-s2) is designated as gc and is used to resolve expressions involving both mass and force expressed as pounds.For instance, in writing Newtons second law, the equation would be written as F = ma/gc, where F is in lbf, m in lbm, and a isin ft/s2.Similar expressions exist for other quantities. Kinetic Energy: KE = mv2/2gc, with KE in (ft-lbf); Potential Energy: PE = mgh/gc,with PE in (ft-lbf); Fluid Pressure: p = ρgh/gc, with p in (lbf/ft2); Specific Weight: SW = ρg/gc, in (lbf/ft3); Shear Stress: τ =(µ/gc)(dv/dy), with shear stress in (lbf/ft2). In all these examples, gc should be regarded as a unit conversion factor. It isfrequently not written explicitly in engineering equations. However, its use is required to produce a consistent set of units.Note that the conversion factor gc [lbm-ft/(lbf-s2)] should not be confused with the local acceleration of gravity g, which hasdifferent units (m/s2) and may be either its standard value (9.807 m/s2) or some other local value.If the problem is presented in USCS units, it may be necessary to use the constant gc in the equation to have a consistent set ofunits. METRIC PREFIXES Multiple Prefix Symbol COMMONLY USED EQUIVALENTS 10–18 atto a 1 gallon of water weighs 8.34 lbf 10–15 femto f 1 cubic foot of water weighs 62.4 lbf 10–12 pico p 1 cubic inch of mercury weighs 0.491 lbf 10–9 nano n The mass of one cubic meter of water is 1,000 kilograms 10–6 micro µ 10–3 milli m 10–2 centi c TEMPERATURE CONVERSIONS 10–1 deci d ºF = 1.8 (ºC) + 32 101 deka da ºC = (ºF – 32)/1.8 102 hecto h ºR = ºF + 459.69 103 kilo k 106 mega K = ºC + 273.15 M 109 giga G 1012 tera T 1015 peta P 1018 exa E FUNDAMENTAL CONSTANTSQuantity Symbol Value Unitselectron charge e 1.6022 × 10−19 C (coulombs)Faraday constant 96,485 coulombs/(mol)gas constant metric R 8,314 J/(kmol·K)gas constant metric R 8.314 kPa·m3/(kmol·K)gas constant USCS R 1,545 ft-lbf/(lb mole-ºR) R 0.08206 L-atm/mole-Kgravitation - newtonian constant G 6.673 × 10–11 m3/(kg·s2)gravitation - newtonian constant G 6.673 × 10–11 N·m2/kg2gravity acceleration (standard) metric g 9.807 m/s2gravity acceleration (standard) USCS g 32.174 ft/s2molar volume (ideal gas), T = 273.15K, p = 101.3 kPa Vm 22,414 L/kmolspeed of light in vacuum c 299,792,000 m/s 1
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CONVERSION FACTORS Multiply By To Obtain Multiply By To Obtainacre 43,560 square feet (ft2) joule (J) 9.478×10–4 Btuampere-hr (A-hr) 3,600 coulomb (C) J 0.7376 ft-lbfångström (Å) 1×10–10 meter (m) J 1 newton·m (N·m)atmosphere (atm) 76.0 cm, mercury (Hg) J/s 1 watt (W)atm, std 29.92 in, mercury (Hg)atm, std 14.70 lbf/in2 abs (psia) kilogram (kg) 2.205 pound (lbm)atm, std 33.90 ft, water kgf 9.8066 newton (N)atm, std 1.013×105 pascal (Pa) kilometer (km) 3,281 feet (ft) km/hr 0.621 mphbar 1×105 Pa kilopascal (kPa) 0.145 lbf/in2 (psi)barrels–oil 42 gallons–oil kilowatt (kW) 1.341 horsepower (hp)Btu 1,055 joule (J) kW 3,413 Btu/hrBtu 2.928×10–4 kilowatt-hr (kWh) kW 737.6 (ft-lbf )/secBtu 778 ft-lbf kW-hour (kWh) 3,413 BtuBtu/hr 3.930×10–4 horsepower (hp) kWh 1.341 hp-hrBtu/hr 0.293 watt (W) kWh 3.6×106 joule (J)Btu/hr 0.216 ft-lbf/sec kip (K) 1,000 lbf K 4,448 newton (N)calorie (g-cal) 3.968×10–3 Btucal 1.560×10–6 hp-hr liter (L) 61.02 in3cal 4.186 joule (J) L 0.264 gal (US Liq)cal/sec 4.186 watt (W) L 10–3 m3centimeter (cm) 3.281×10–2 foot (ft) L/second (L/s) 2.119 ft3/min (cfm)cm 0.394 inch (in) L/s 15.85 gal (US)/min (gpm)centipoise (cP) 0.001 pascal·sec (Pa·s)centistokes (cSt) 1×10–6 m2/sec (m2/s) meter (m) 3.281 feet (ft)cubic feet/second (cfs) 0.646317 million gallons/day m 1.094 yard (mgd)cubic foot (ft3) 7.481 gallon m/second (m/s) 196.8 feet/min (ft/min)cubic meters (m3) 1,000 Liters mile (statute) 5,280 feet (ft)electronvolt (eV) 1.602×10–19 joule (J) mile (statute) 1.609 kilometer (km) mile/hour (mph) 88.0 ft/min (fpm)foot (ft) 30.48 cm mph 1.609 km/hft 0.3048 meter (m) mm of Hg 1.316×10–3 atm –3ft-pound (ft-lbf) 1.285×10 Btu mm of H2O 9.678×10–5 atmft-lbf 3.766×10–7 kilowatt-hr (kWh)ft-lbf 0.324 calorie (g-cal) newton (N) 0.225 lbfft-lbf 1.356 joule (J) N·m 0.7376 ft-lbfft-lbf/sec 1.818×10–3 horsepower (hp) N·m 1 joule (J)gallon (US Liq) 3.785 liter (L) pascal (Pa) 9.869×10–6 atmosphere (atm)gallon (US Liq) 0.134 ft3 Pa 1 newton/m2 (N/m2)gallons of water 8.3453 pounds of water Pa·sec (Pa·s) 10 poise (P)gamma (γ, Γ) 1×10–9 tesla (T) pound (lbm,avdp) 0.454 kilogram (kg)gauss 1×10–4 T lbf 4.448 Ngram (g) 2.205×10–3 pound (lbm) lbf-ft 1.356 N·m lbf/in2 (psi) 0.068 atmhectare 1×104 square meters (m2) psi 2.307 ft of H2Ohectare 2.47104 acres psi 2.036 in of Hghorsepower (hp) 42.4 Btu/min psi 6,895 Pahp 745.7 watt (W)hp 33,000 (ft-lbf)/min radian 180/π degreehp 550 (ft-lbf)/sechp-hr 2,544 Btu stokes 1×10–4 m2/s 6hp-hr 1.98×10 ft-lbfhp-hr 2.68×106 joule (J) therm 1×105 Btuhp-hr 0.746 kWh watt (W) 3.413 Btu/hrinch (in) 2.540 centimeter (cm) W 1.341×10–3 horsepower (hp)in of Hg 0.0334 atm W 1 joule/sec (J/s)in of Hg 13.60 in of H2O weber/m2 (Wb/m2) 10,000 gaussin of H2O 0.0361 lbf/in2 (psi)in of H2O 0.002458 atm 2
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MATHEMATICS (continued)Case 4. Circle e = 0: h = –a; k = –b (x – h)2 + (y – k)2 = r2; Center at (h, k) r = a 2 + b2 − cis the general form of the equation with radius If a2 + b2 – c is positive, a circle, center (–a, –b).r= (x − h )2 + ( y − k )2 If a2 + b2 – c equals zero, a point at (–a, –b). If a2 + b2 – c is negative, locus is imaginary.• QUADRIC SURFACE (SPHERE) The general form of the equation is (x – h)2 + (y – k)2 + (z – m)2 = r2 with center at (h, k, m). In a three-dimensional space, the distance between two points is d= (x2 − x1 )2 + ( y 2 − y1 )2 + (z 2 − z1 )2Length of the tangent from a point. Using the general formof the equation of a circle, the length of the tangent is found LOGARITHMSfrom The logarithm of x to the Base b is defined by t2 = (x′ – h)2 + (y′ – k)2 – r2 logb (x) = c, where bc = xby substituting the coordinates of a point P(x′,y′) and the Special definitions for b = e or b = 10 are:coordinates of the center of the circle into the equation and ln x, Base = ecomputing. log x, Base = 10 To change from one Base to another:• logb x = (loga x)/(loga b) e.g., ln x = (log10 x)/(log10 e) = 2.302585 (log10 x) Identities logb bn = n log xc = c log x; xc = antilog (c log x) log xy = log x + log y logb b = 1; log 1 = 0 log x/y = log x – log yConic Section EquationThe general form of the conic section equation is • Brink, R.W., A First Year of College Mathematics, Copyright 1937 by D. Ax2 + 2Bxy + Cy2 + 2Dx + 2Ey + F = 0 Appleton-Century Co., Inc. Reprinted by permission of Prentice-Hall, Inc., Englewood Cliffs, NJ. where not both A and C are zero. 2If B – AC < 0, an ellipse is defined.If B2 – AC > 0, a hyperbola is defined.If B2 – AC = 0, the conic is a parabola.If A = C and B = 0, a circle is defined.If A = B = C = 0, a straight line is defined. x2 + y2 + 2ax + 2by + c = 0is the normal form of the conic section equation, if thatconic section has a principal axis parallel to a coordinateaxis. 4
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MATHEMATICS (continued)TRIGONOMETRY sin α sin β = (1/2)[cos (α – β) – cos (α + β)]Trigonometric functions are defined using a right triangle. cos α cos β = (1/2)[cos (α – β) + cos (α + β)]sin θ = y/r, cos θ = x/r sin α cos β = (1/2)[sin (α + β) + sin (α – β)]tan θ = y/x, cot θ = x/y sin α + sin β = 2 sin (1/2)(α + β) cos (1/2)(α – β)csc θ = r/y, sec θ = r/x sin α – sin β = 2 cos (1/2)(α + β) sin (1/2)(α – β) cos α + cos β = 2 cos (1/2)(α + β) cos (1/2)(α – β) cos α – cos β = – 2 sin (1/2)(α + β) sin (1/2)(α – β) a b c COMPLEX NUMBERS Law of Sines = = sin A sin B sin C Definition i = −1 Law of Cosines (a + ib) + (c + id) = (a + c) + i (b + d) 2 2 2 a = b + c – 2bc cos A (a + ib) – (c + id) = (a – c) + i (b – d) 2 2 2 b = a + c – 2ac cos B (a + ib)(c + id) = (ac – bd) + i (ad + bc) a + ib (a + ib )(c − id ) (ac + bd ) + i(bc − ad ) 2 2 2 c = a + b – 2ab cos C = =Identities c + id (c + id )(c − id ) c2 + d 2csc θ = 1/sin θ (a + ib) + (a – ib) = 2asec θ = 1/cos θ (a + ib) – (a – ib) = 2ibtan θ = sin θ/cos θ (a + ib)(a – ib) = a2 + b2cot θ = 1/tan θ Polar Coordinates 2 2sin θ + cos θ = 1 x = r cos θ; y = r sin θ; θ = arctan (y/x)tan2θ + 1 = sec2θ r = x + iy = x2 + y2cot2θ + 1 = csc2θ x + iy = r (cos θ + i sin θ) = reiθsin (α + β) = sin α cos β + cos α sin β [r1(cos θ1 + i sin θ1)][r2(cos θ2 + i sin θ2)] =cos (α + β) = cos α cos β – sin α sin β r1r2[cos (θ1 + θ2) + i sin (θ1 + θ2)]sin 2α = 2 sin α cos α n 2 2 2 2 (x + iy) = [r (cos θ + i sin θ)]ncos 2α = cos α – sin α = 1 – 2 sin α = 2 cos α – 1 2 = r n(cos nθ + i sin nθ)tan 2α = (2 tan α)/(1 – tan α) r1 (cos θ + i sin θ1 ) rcot 2α = (cot2α – 1)/(2 cot α) = 1 [cos(θ1 − θ 2 ) + i sin (θ1 − θ 2 )] r2 (cos θ 2 + i sin θ 2 ) r2tan (α + β) = (tan α + tan β)/(1 – tan α tan β) Eulers Identitycot (α + β) = (cot α cot β – 1)/(cot α + cot β) eiθ = cos θ + i sin θsin (α – β) = sin α cos β – cos α sin β e−iθ = cos θ – i sin θcos (α – β) = cos α cos β + sin α sin β eiθ + e −iθ eiθ − e −iθtan (α – β) = (tan α – tan β)/(1 + tan α tan β) cos θ = , sin θ = 2 2icot (α – β) = (cot α cot β + 1)/(cot β – cot α) Rootssin (α/2) = ± (1 − cos α ) 2 If k is any positive integer, any complex number (other than zero) has k distinct roots. The k roots of r (cos θ + i sin θ)cos (α/2) = ± (1 + cos α ) 2 can be found by substituting successively n = 0, 1, 2, …,tan (α/2) = ± (1 − cos α ) (1 + cos α ) (k – 1) in the formulacot (α/2) = ± (1 + cos α ) (1 − cos α ) é æθ 360o ö æθ 360o öù w = k r êcosç + n ç ÷ + i sin ç + n ÷ú ê èk ë k ÷ ø çk è k ÷úøû 5
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MATHEMATICS (continued)MATRICES For a third-order determinant:A matrix is an ordered rectangular array of numbers with mrows and n columns. The element aij refers to row i and a1 a2 a3column j. b1 b2 b3 = a1b2 c3 + a2 b3 c1 + a3b1c2 − a3b2 c1 − a2b1c3 − a1b3c2Multiplication c1 c2 c3If A = (aik) is an m × n matrix and B = (bkj) is an n × smatrix, the matrix product AB is an m × s matrix VECTORS ( ) æn ö C = ci j = ç å ail blj ÷ è l =1 ø jwhere n is the common integer representing the number ofcolumns of A and the number of rows of B (l and k = 1, 2,…, n). i kAdditionIf A = (aij) and B = (bij) are two matrices of the same sizem × n, the sum A + B is the m × n matrix C = (cij) wherecij = aij + bij.IdentityThe matrix I = (aij) is a square n × n identity matrix whereaii = 1 for i = 1, 2, …, n and aij = 0 for i ≠ j.TransposeThe matrix B is the transpose of the matrix A if each entrybji in B is the same as the entry aij in A and conversely. In A = axi + ayj + azkequation form, the transpose is B = AT. Addition and subtraction:Inverse A + B = (ax + bx)i + (ay + by)j + (az + bz)kThe inverse B of a square n × n matrix A is adj ( A) A – B = (ax – bx)i + (ay – by)j + (az – bz)k B = A −1 = , where A The dot product is a scalar product and represents the projection of B onto A times A. It is given byadj(A) = adjoint of A (obtained by replacing AT elements A·B = axbx + ayby + azbzwith their cofactors, see DETERMINANTS) and = AB cos θ = B·AA = determinant of A. The cross product is a vector product of magnitudeDETERMINANTS BA sin θ which is perpendicular to the planeA determinant of order n consists of n2 numbers, called the containing A and B. The product iselements of the determinant, arranged in n rows and ncolumns and enclosed by two vertical lines. In any i j kdeterminant, the minor of a given element is the A × B = a x a y a z = −B × Adeterminant that remains after all of the elements are struckout that lie in the same row and in the same column as the bx b y bzgiven element. Consider an element which lies in the hthcolumn and the kth row. The cofactor of this element is thevalue of the minor of the element (if h + k is even), and it isthe negative of the value of the minor of the element (if h +k is odd).If n is greater than 1, the value of a determinant of order nis the sum of the n products formed by multiplying eachelement of some specified row (or column) by its cofactor.This sum is called the expansion of the determinant[according to the elements of the specified row (or The sense of A × B is determined by the right-hand rule.column)]. For a second-order determinant: A × B = AB n sin θ, where a1 a2 = a1b2 − a2b1 n = unit vector perpendicular to the plane of A and B. b1 b2 6
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MATHEMATICS (continued)Gradient, Divergence, and Curl Geometric Progression æ ∂ ∂ ∂ ö To determine whether a given finite sequence is a∇φ = ç i + ç ∂x j + k ÷φ geometric progression (G.P.), divide each number after the è ∂y ∂z ÷ ø first by the preceding number. If the quotients are equal, the series is geometric. æ ∂∇⋅V = ç i + ç ∂x ∂ ∂y ∂ ö ( j + k ÷ ⋅ V1 i + V2 j + V3k ∂z ÷ ) 1. The first term is a. è ø 2. The common ratio is r. æ ∂∇×V = ç i + ç ∂x ∂ ∂y ∂ ö ( j + k ÷ × V1 i + V2 j + V3k ∂z ÷ ) 3. The number of terms is n. è ø 4. The last or nth term is l.The Laplacian of a scalar function φ is 5. The sum of n terms is S. 2 ∂ 2φ ∂ 2φ ∂ 2φ∇ φ= 2 + 2 + 2 l = arn−1 ∂x ∂y ∂z S = a (1 – rn)/(1 – r); r ≠ 1Identities S = (a – rl)/(1 – r); r ≠ 1A·B = B·A; A·(B + C) = A·B + A·C limit S n = a (1 − r ); r < 1A·A = A2 n →∞i·i = j·j = k·k = 1 A G.P. converges if r < 1 and it diverges if r ≥ 1.i·j = j·k = k·i = 0 Properties of SeriesIf A·B = 0, then either A = 0, B = 0, or A is perpendicular nto B. å c = nc; c = constant i =1A × B = –B × A n n å cxi = c å xiA × (B + C) = (A × B) + (A × C) i =1 i =1(B + C) × A = (B × A) + (C × A) n n n å (xi + yi − zi ) = å xi + å yi − å zi n i =1 i =1 i =1 i =1i×i =j×j =k×k=0i × j = k = –j × i; j × k = i = –k × j n x =1 ( 2 åx= n+n 2 )k × i = j = –i × k 1. A power series in x, or in x – a, which is convergent inIf A × B = 0, then either A = 0, B = 0, or A is parallel to B. the interval –1 < x < 1 (or –1 < x – a < 1), defines a function of x which is continuous for all values of x∇ 2 φ = ∇ ⋅ (∇ φ) = (∇ ⋅ ∇ )φ within the interval and is said to represent the function∇ × ∇φ = 0 in that interval.∇ ⋅ (∇ × A ) = 0 2. A power series may be differentiated term by term, and the resulting series has the same interval of∇ × (∇ × A) = ∇ (∇ ⋅ A) − ∇ 2 A convergence as the original series (except possibly at the end points of the interval).PROGRESSIONS AND SERIES 3. A power series may be integrated term by termArithmetic Progression provided the limits of integration are within the intervalTo determine whether a given finite sequence of numbers is of convergence of the series.an arithmetic progression, subtract each number from the 4. Two power series may be added, subtracted, orfollowing number. If the differences are equal, the series is multiplied, and the resulting series in each case isarithmetic. convergent, at least, in the interval common to the two1. The first term is a. series.2. The common difference is d. 5. Using the process of long division (as for polynomials), two power series may be divided one by the other.3. The number of terms is n.4. The last or nth term is l.5. The sum of n terms is S. l = a + (n – 1)d S = n(a + l)/2 = n [2a + (n – 1) d]/2 7
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MATHEMATICS (continued)Taylors Series Property 3. Law of Compound or Joint Probability f ′(a ) (x − a ) + f (a ) (x − a )2 ′′ If neither P(A) nor P(B) is zero, f (x ) = f (a ) + 1! 2! P(A, B) = P(A)P(B | A) = P(B)P(A | B), where f (n ) (a ) P(B | A) = the probability that B occurs given the fact that A +K+ (x − a )n + K has occurred, and n!is called Taylors series, and the function f (x) is said to be P(A | B) = the probability that A occurs given the fact that Bexpanded about the point a in a Taylors series. has occurred.If a = 0, the Taylors series equation becomes a Maclaurins If either P(A) or P(B) is zero, then P(A, B) = 0.series. Probability FunctionsPROBABILITY AND STATISTICS A random variable x has a probability associated with each of its values. The probability is termed a discretePermutations and Combinations probability if x can assume only the discrete valuesA permutation is a particular sequence of a given set of x = X1, X2, …, Xi, …, XNobjects. A combination is the set itself without reference toorder. The discrete probability of the event X = xi occurring is defined as P(Xi).1. The number of different permutations of n distinct objects taken r at a time is Probability Density Functions n! If x is continuous, then the probability density function f (x) P (n , r ) = (n − r ) ! is defined so that òx f ( x )dx = the probability that x lies between x1 and x2. x 22. The number of different combinations of n distinct 1 objects taken r at a time is The probability is determined by defining the equation for f (x) and integrating between the values of x required. P(n , r ) n! C (n , r ) = = r! [r!(n − r )!] Probability Distribution Functions The probability distribution function F(Xn) of the discrete3. The number of different permutations of n objects probability function P(Xi) is defined by taken n at a time, given that ni are of type i, where i = 1, 2,…, k and Σni = n, is n F ( X n ) = å P ( X k ) = P( X i ≤ X n ) k =1 n! P (n; n1 , n2 ,K nk ) = n1!n2 !Kn k ! When x is continuous, the probability distribution function F(x) is defined byLaws of Probability F (x ) = ò−∞ f (t )dt xProperty 1. General Character of ProbabilityThe probability P(E) of an event E is a real number in the which implies that F(a) is the probability that x ≤ a.range of 0 to 1. The probability of an impossible event is 0 The expected value g(x) of any function is defined asand that of an event certain to occur is 1. E{g (x )} = ò −∞ g (t ) f (t )dt xProperty 2. Law of Total Probability P(A + B) = P(A) + P(B) – P(A, B), where Binomial DistributionP(A + B) = the probability that either A or B occur alone P(x) is the probability that x will occur in n trials. If p = or that both occur together, probability of success and q = probability of failure = 1 – p,P(A) = the probability that A occurs, then n!P(B) = the probability that B occurs, and P (x ) = C (n , x ) p x q n − x = p x q n − x , where x! (n − x ) !P(A, B) = the probability that both A and B occur simultaneously. x = 0, 1, 2, …, n, C(n, x) = the number of combinations, and n, p = parameters. 8
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MATHEMATICS (continued)Normal Distribution (Gaussian Distribution) The standard deviation of a population isThis is a unimodal distribution, the mode being x = µ, with σ= (1 N )å ( X i − µ )2two points of inflection (each located at a distance σ toeither side of the mode). The averages of n observations The sample variance istend to become normally distributed as n increases. The 2 ( ) nvariate x is said to be normally distributed if its density s 2 = [1 (n − 1)]å X i − Xfunction f (x) is given by an expression of the form i =1 1 The sample standard deviation is f (x ) = e − ( x −µ ) 2 2σ 2 , where σ 2πµ = the population mean, é 1 ùn s= ê úå Xi − X ( )2 ë n − 1û i =1σ = the standard deviation of the population, and The coefficient of variation = CV = s/ X–∞ ≤ x ≤ ∞When µ = 0 and σ2 = σ = 1, the distribution is called a The geometric mean = n X1 X 2 X 3 K X nstandardized or unit normal distribution. Then The root-mean-square value = (1 n)å X i2 1 f (x ) = 2 e−x 2 , where − ∞ ≤ x ≤ ∞ . 2π The median is defined as the value of the middle item when the data are rank-ordered and the number of items is odd.A unit normal distribution table is included at the end of The median is the average of the middle two items whenthis section. In the table, the following notations are the rank-ordered data consists of an even number of items.utilized: The mode of a set of data is the value that occurs withF(x) = the area under the curve from –∞ to x, greatest frequency.R(x) = the area under the curve from x to ∞, and t-DistributionW(x) = the area under the curve between –x and x. The variate t is defined as the quotient of two independentDispersion, Mean, Median, and Mode Values variates x and r where x is unit normal and r is the rootIf X1, X2, …, Xn represent the values of n items or mean square of n other independent unit normal variates;observations, the arithmetic mean of these items or that is,observations, denoted X , is defined as t = x/r. The following is the t-distribution with n degrees of freedom: n X = (1 n )( X 1 + X 2 + K + X n ) = (1 n )å X i Γ [(n + 1)] 2 1 f (t ) = ( )( i =1 n +1) 2 Γ(n 2) nπ 1 + t 2 n X → µ for sufficiently large values of n. where – ∞ ≤ t ≤ ∞.The weighted arithmetic mean is A table at the end of this section gives the values of tα, n for å wi X i Xw = , where values of α and n. Note that in view of the symmetry of the å wi t-distribution,X w = the weighted arithmetic mean, t1−α,n = –tα,n. The function for α follows:Xi = the values of the observations to be averaged, and α = òt∞,n f (t )dtwi = the weight applied to the Xi value. αThe variance of the observations is the arithmetic mean of A table showing probability and density functions isthe squared deviations from the population mean. In included on page 149 in the INDUSTRIALsymbols, X1, X2, …, Xn represent the values of the n sample ENGINEERING SECTION of this handbook.observations of a population of size N. If µ is the arithmeticmean of the population, the population variance is definedby σ 2 = (1 / N )[( X 1 − µ ) + ( X 2 − µ ) + K + ( X N − µ ) ] 2 2 2 2 = (1 / N )å ( X i − µ ) N i =1 9
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MATHEMATICS (continued)GAMMA FUNCTION Γ (n ) = òo t n −1e −t dt , n > 0 ∞CONFIDENCE INTERVALSConfidence Interval for the Mean µ of a NormalDistribution(a) Standard deviation σ is known σ σ X − Zα 2 ≤ µ ≤ X + Zα 2 n n(b) Standard deviation σ is not known s s X − tα 2 ≤ µ ≤ X + tα 2 n nwhere t α 2 corresponds to n – 1 degrees of freedom.Confidence Interval for the Difference Between TwoMeans µ1 and µ2(a) Standard deviations σ1 and σ2 known σ1 σ 2 2 σ2 σ2 X1 − X 2 − Zα 2 + 2 ≤ µ1 − µ 2 ≤ X 1 − X 2 + Z α 2 1 + 2 n1 n2 n1 n2(b) Standard deviations σ1 and σ2 are not known æ 1 ç 1 ö ÷[ ç n + n ÷ (n − 1)S 1 + (n 2 − 1)S 2 2 2 ] æ 1 1 ö [ ç n + n ÷ (n − 1)S 1 + (n 2 − 1)S 2 ç ÷ 2 2 ] è 1 2 ø è 1 2 ø X 1 − X 2 − tα 2 ≤ µ1 − µ 2 ≤ X 1 − X 2 − t α 2 n1 + n 2 − 2 n1 + n 2 − 2where t α 2 corresponds to n1 + n2 – 2 degrees of freedom. 10
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MATHEMATICS (continued) DERIVATIVES AND INDEFINITE INTEGRALSIn these formulas, u, v, and w represent functions of x. Also, a, c, and n represent constants. All arguments of the trigonometricfunctions are in radians. A constant of integration should be added to the integrals. To avoid terminology difficulty, thefollowing definitions are followed: arcsin u = sin–1 u, (sin u) –1 = 1/sin u.1. dc/dx = 0 1. ò d f (x) = f (x)2. dx/dx = 1 2. ò dx = x3. d(cu)/dx = c du/dx 3. ò a f(x) dx = a ò f(x) dx4. d(u + v – w)/dx = du/dx + dv/dx – dw/dx 4. ò [u(x) ± v(x)] dx = ò u(x) dx ± ò v(x) dx5. d(uv)/dx = u dv/dx + v du/dx x m+16. d(uvw)/dx = uv dw/dx + uw dv/dx + vw du/dx 5. ò x m dx = (m ≠ −1) m +1 d (u v ) v du dx − u dv dx 6. ò u(x) dv(x) = u(x) v(x) – ò v (x) du(x)7. = dx v2 dx 1 n n–1 7. ò = ln ax + b8. d(u )/dx = nu du/dx ax + b a9. d[f (u)]/dx = {d[f (u)]/du} du/dx dx10. du/dx = 1/(dx/du) 8. ò =2 x x d (log a u ) 1 du11. = (log a e ) ax dx u dx 9. ò ax dx = ln a d (lnu ) 1 du12. = 10. ò sin x dx = – cos x dx u dx 11. ò cos x dx = sin x13. ( ) d au = (lna )a u du 12. ò sin 2 xdx = x sin 2 x − dx dx 2 414. d(eu)/dx = eu du/dx x sin 2 x15. d(uv)/dx = vuv–1 du/dx + (ln u) uv dv/dx 13. ò cos 2 xdx = + 2 416. d(sin u)/dx = cos u du/dx 14. ò x sin x dx = sin x – x cos x17. d(cos u)/dx = –sin u du/dx 15. ò x cos x dx = cos x + x sin x18. d(tan u)/dx = sec2u du/dx 16. ò sin x cos x dx = (sin2x)/219. d(cot u)/dx = –csc2u du/dx cos(a − b )x cos(a + b )x20. d(sec u)/dx = sec u tan u du/dx 17. ò sin ax cos bx dx = − 2(a − b ) − 2(a + b ) (a 2 ≠ b2 )21. d(csc u)/dx = –csc u cot u du/dx 18. ò tan x dx = –lncos x = ln sec x22. ( d sin −1u = 1) du (− π 2 ≤ sin −1 u≤π 2 ) 19. ò cot x dx = –ln csc x = ln sin x dx 1 − u 2 dx 20. ò tan2x dx = tan x – x 21. ò cot2x dx = –cot x – x23. ( d cos −1u =− ) 1 du (0 ≤ cos −1 u≤π ) 22. ò eax dx = (1/a) eax dx 2 dx 23. ò xeax dx = (eax/a2)(ax – 1) 1− u 24. ò ln x dx = x [ln (x) – 1] (x > 0)24. ( d tan u = −1 ) 1 du (− π 2 < tan −1 u<π 2 ) 25. ò 2 dx 2 1 = tan −1 x (a ≠ 0) dx 1 + u 2 dx a +x a a æ aö25. ( d cot −1u =− )1 du (0 < cot −1 u<π ) 26. ò dx 2 ax + c = 1 ac tan −1 ç x ç c÷ ÷, (a > 0, c > 0) è ø dx 1 + u 2 dx dx 2 2ax + b = tan −1 ( d sec −1u = ) 1 du ò 2 27a. ax + bx + c 4ac − b 2 4ac − b 226. dx u u − 1 dx 2 (4ac − b 2 >0 ) (0 ≤ sec −1 )( −1 u < π 2 − π ≤ sec u < − π 2 ) dx 1 2ax + b − b 2 − 4ac ò = ln ( d csc −1u =− 1 ) du 2 27b. ax + bx + c b 2 − 4ac 2ax + b + b 2 − 4ac27. dx u u 2 − 1 dx (b 2 − 4ac > 0 ) (0 < csc −1 )( u ≤ π 2 − π < csc −1u ≤ − π 2 ) 27c. ò dx =− 2 , (b 2 − 4ac = 0 ) 2 ax + bx + c 2ax + b 15
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MATHEMATICS (continued) MENSURATION OF AREAS AND VOLUMESNomenclature Circular SegmentA = total surface area ♦P = perimeterV = volumeParabola A = [r2 (φ – sin φ)]/2 φ = s/r = 2{arccos [(r – d)/r]} Circular Sector ♦Ellipse♦ A = φr2/2 = sr/2 φ = s/r Sphere ♦ A = πab (Papprox = 2π a 2 + b 2 2 ) é1 + ( 1 2 )2 λ2 + ( 1 2 × 1 4 )2 λ4 ù ê úP = π(a + b )ê+ ( 1 × 1 × 3 )2 λ6 + ( 1 × 1 × 3 × 5 )2 λ8 ú , 2 4 6 2 4 6 8 ê ú ê+ ( 1 × 1 × 3 × 5 × 7 )2 λ10 + K ú ê ë 2 4 6 8 10 ú ûwhere V = 4πr3/3 = πd 3/6λ = (a – b)/(a + b) A = 4πr2 = πd 2 ♦ Gieck, K. & Gieck R., Engineering Formulas, 6th Ed., Copyright 1967 by Gieck Publishing. Diagrams reprinted by permission of Kurt Gieck. 16
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