ALGEBRA 1 Man-hours (is always assumed constant) LOGARITHM (Wor ker s1 )(time1 ) (Wor ker s 2 )(time2 ) = quantity.of .work1 quantity.of .work 2 x = log b N → N = b x Properties ALGEBRA 2 log( xy ) = log x + log y x log = log x − log y y log x n = n log x log x log b x = log b log a a = 1 UNIFORM MOTION PROBLEMS REMAINDER AND FACTOR THEOREMS Traveling against the wind or upstream: S = Vt Traveling with the wind or downstream: Vtotal = V1 + V2 Given: f ( x) (x − r) Vtotal = V1 − V2 DIGIT AND NUMBER PROBLEMS Remainder Theorem: Remainder = f(r) Factor Theorem: Remainder = zero 100h + 10t + u → QUADRATIC EQUATIONS where: Ax 2 + Bx + C = 0 Root = − B ± B 2 − 4 AC 2A 2-digit number h = hundred’s digit t = ten’s digit u = unit’s digit CLOCK PROBLEMS Sum of the roots = - B/A Products of roots = C/A MIXTURE PROBLEMS Quantity Analysis: A + B = C Composition Analysis: Ax + By = Cz where: WORK PROBLEMS Rate of doing work = 1/ time Rate x time = 1 (for a complete job) Combined rate = sum of individual rates PDF created with pdfFactory trial version www.pdffactory.com x = distance traveled by the minute hand in minutes x/12 = distance traveled by the hour hand in minutes PROGRESSION PROBLEMS a1 an am d S = = = = = HARMONIC PROGRESSION (HP) first term nth term any term before an common difference sum of all “n” terms - Mean – middle term or terms between two terms in the progression. ARITHMETIC PROGRESSION (AP) - difference of any 2 no.’s is constant calcu function: LINEAR (LIN) a n = a m + ( n − m) d nth term d = a2 − a1 = a3 − a2 ,...etc S= n (a1 + an ) 2 S= n [2 a1 + (n − 1) d ] 2 Common difference Sum of ALL terms RATIO of any 2 adj, terms is always constant Calcu function: EXPONENTIAL (EXP) an = a m r n−m COIN PROBLEMS Penny = 1 centavo coin Nickel = 5 centavo coin Dime = 10 centavo coin Quarter = 25 centavo coin Half-Dollar = 50 centavo coin DIOPHANTINE EQUATIONS If the number of equations is less than the number of unknowns, then the equations are called “Diophantine Equations”. Sum of ALL terms GEOMETRIC PROGRESSION (GP) - a sequence of number in which their reciprocals form an AP calcu function: LINEAR (LIN) nth term ALGEBRA 3 Fundamental Principle: “If one event can occur in m different ways, and after it has occurred in any one of these ways, a second event can occur in n different ways, and then the number of ways the two events can occur in succession is mn different ways” PERMUTATION Permutation of n objects taken r at a time nPr = r= a 2 a3 = a1 a2 n! (n − r )! ratio a ( r n − 1) S= 1 → r >1 r −1 a1 (1 − r n ) S= → r <1 1− r Permutation of n objects taken n at a time Sum of ALL terms, r >1 nPn = n! Permutation of n objects with q,r,s, etc. objects are alike P= Sum of ALL terms, r < 1 n! q!r!s!... Permutation of n objects arrange in a circle S= a1 1− r → r < 1& n = ∞ Sum of ALL terms, r<1,n=∞ PDF created with pdfFactory trial version www.pdffactory.com P = ( n − 1)! COMBINATION PROBABILITY Combination of n objects taken r at a time Probability of an event to occur (P) nCr = n! (n − r )!r! Combination of n objects taken n at a time P= number _ of _ successful _ outcomes total _ outcomes Probability of an event not to occur (Q) nCn = 1 Q=1–P Combination of n objects taken 1, 2, 3…n at a time C = 2n − 1 MULTIPLE EVENTS Mutually exclusive events without a common outcome PA or B = PA + PB BINOMIAL EXPANSION Properties of a binomial expansion: (x + y)n Mutually exclusive events with a common outcome 1. The number of terms in the resulting expansion is PA or B = PA + PB – PA&B equal to “n+1” 2. The powers of x decreases by 1 in the successive terms while the powers of y increases Dependent/Independent Probability by 1 in the successive terms. PAandB =PA × PB 3. The sum of the powers in each term is always equal to “n” 4. The first term is xn while the last term in yn both of the terms having a coefficient of 1. th r term in the expansion (x + y) term involving yr in the expansion (x + y)n y term = nCr (x) n-r (y) P = nCr pr qn-r n r th term = nCr-1 (x)n-r+1 (y)r-1 r REPEATED TRIAL PROBABILITY r sum of coefficients of (x + y)n Sum = (coeff. of x + coeff. of y) n p = probability that the event happen q = probability that the event failed VENN DIAGRAMS Venn diagram in mathematics is a diagram representing a set or sets and the logical, relationships between them. The sets are drawn as circles. The method is named after the British mathematician and logician John Venn. sum of coefficients of (x + k)n Sum = (coeff. of x + k)n – (k)n PDF created with pdfFactory trial version www.pdffactory.com PLANE TRIGONOMETRY ANGLE, MEASUREMENTS & CONVERSIONS 1 revolution = 360 degrees 1 revolution = 2π radians 1 revolution = 400 grads 1 revolution = 6400 mils 1 revolution = 6400 gons Relations between two angles (A & B) Complementary angles → A + B = 90° Supplementary angles → A + B = 180° Explementary angles → A + B = 360° TRIGONOMETRIC IDENTITIES sin 2 A + cos2 A = 1 1 + cot 2 A = csc 2 A 1 + tan 2 A = sec 2 A sin( A ± B) = sin A cos B ± cos A sin B cos( A ± B) = cos A cos B m sin A sin B tan A ± tan B tan( A ± B) = 1 m tan A tan B cot A cot B m 1 cot( A ± B) = cot A ± cot B sin 2 A = 2 sin A cos B cos 2 A = cos 2 A − sin 2 A 2 tan A tan 2 A = 1 − tan 2 A cot 2 A − 1 cot 2 A = 2 cot A SOLUTIONS TO OBLIQUE TRIANGLES SINE LAW Angle (θ) REFLEX Measurement θ = 0° 0° < θ < 90° θ = 90° 90° < θ < 180° θ =180° 180° < θ < 360° FULL OR PERIGON θ = 360° NULL ACUTE RIGHT OBTUSE STRAIGHT a b c = = sin A sin B sin C COSINE LAW a2 = b2 + c2 – 2 b c cos A b2 = a2 + c2 – 2 a c cos B c2 = a2 + b2 – 2 a b cos C Pentagram – golden triangle (isosceles) 36° AREAS OF TRIANGLES AND QUADRILATERALS 72° 72° TRIANGLES 1. Given the base and height Area = 1 bh 2 2. Given two sides and included angle Area = PDF created with pdfFactory trial version www.pdffactory.com 1 ab sin θ 2 3. Given three sides 4. Quadrilateral circumscribing in a circle Area = s ( s − a )(s − b)( s − c) s= Area = rs a+b+c 2 Area = abcd a+b+c+d s= 2 4. Triangle inscribed in a circle Area = abc 4r THEOREMS IN CIRCLES 5. Triangle circumscribing a circle Area = rs 6. Triangle escribed in a circle Area = r ( s − a ) QUADRILATERALS 1. Given diagonals and included angle Area = 1 d1d 2 sin θ 2 2. Given four sides and sum of opposite angles Area = ( s − a)( s − b)(s − c )(s − d ) − abcd cos 2 θ A+C B + D = 2 2 a+b+c+d s= 2 θ= 3. Cyclic quadrilateral – is a quadrilateral inscribed in a circle Area = ( s − a)( s − b)(s − c )(s − d ) a+b+c+d 2 (ab + cd )(ac + bd )(ad + bc ) r= 4( Area ) s= d1 d 2= ac + bd → Ptolemy’s Theorem PDF created with pdfFactory trial version www.pdffactory.com SIMILAR TRIANGLES 2 2 Number of diagonal lines (N): 2 A1 A B C H = = = = A2 a b c h N= 2 SOLID GEOMETRY Area of a regular polygon inscribed in a circle of radius r Area = POLYGONS 3 sides – Triangle 4 sides – Quadrilateral/Tetragon/Quadrangle 5 sides – Pentagon 6 sides – Hexagon 7 sides – Heptagon/Septagon 8 sides – Octagon 9 sides – Nonagon/Enneagon 10 sides – Decagon 11 sides – Undecagon 12 sides – Dodecagon 15 sides – Quidecagon/ Pentadecagon 16 sides – Hexadecagon 20 sides – Icosagon 1000 sides – Chillagon 180° Area = nr 2 tan n Area of a regular polygon having each side measuring x unit length Area = 1 2 180° nx cot 4 n PLANE GEOMETRIC FIGURES CIRCLES πd 2 = πr 2 4 Circumference = πd = 2πr A= Sector of a Circle Sum of interior angles: A= S = n θ = (n – 2) 180° A= Value of each interior angle ( n − 2)(180°) n Value of each exterior angle α = 180° − θ = 1 2 360° nr sin 2 n Area of a regular polygon circumscribing a circle of radius r Let: n = number of sides θ = interior angle α = exterior angle θ= n ( n − 3) 2 360° n 360° s = rθ ( rad ) = πrθ (deg) 180° Segment of a Circle A segment = A sector – A triangle ELLIPSE A=πab Sum of exterior angles: S = n α = 360° 1 1 rs = r 2θ 2 2 2 πr θ (deg) PARABOLIC SEGMENT A= PDF created with pdfFactory trial version www.pdffactory.com 2 bh 3 PYRAMID TRAPEZOID 1 A = ( a + b) h 2 PARALLELOGRAM A = ab sin α A = bh 1 A = d1d 2 sin θ 2 1 Bh 3 A( lateral ) = ∑ Afaces V = A( surface) = A( lateral ) + B Frustum of a Pyramid V = RHOMBUS 1 A = d1d 2 = ah 2 A = a 2 sin α h ( A1 + A2 + A1 A2 ) 3 A1 = area of the lower base A2 = area of the upper base PRISMATOID V = SOLIDS WITH PLANE SURFACE h ( A1 + A2 + 4 Am ) 6 Lateral Area = (No. of Faces) (Area of 1 Face) Am = area of the middle section Polyhedron – a solid bounded by planes. The bounding planes are referred to as the faces and the intersections of the faces are called the edges. The intersections of the edges are called vertices. REGULAR POLYHEDRON PRISM V = Bh a solid bounded by planes whose faces are congruent regular polygons. There are five regular polyhedrons namely: A. B. C. D. E. A(lateral) = PL A(surface) = A(lateral) + 2B where: P = perimeter of the base L = slant height B = base area Truncated Prism ∑ heights V = B number of heights PDF created with pdfFactory trial version www.pdffactory.com Tetrahedron Hexahedron (Cube) Octahedron Dodecahedron Icosahedron Octahedron Triangle Icosahedron Hexahedron Square Triangle Tetrahedron Triangle Pentagon Dodecahedron Name Type of FACE h ( A1 + A2 + A1 A2 3 A( lateral ) = π ( R + r ) L V = SPHERES AND ITS FAMILIES No. of FACES 4 6 8 12 20 6 12 12 30 30 SPHERE No. of EDGES 4 3 πr 3 A( surface ) = 4πr 2 V = 12 V = 2.18 x 3 20 V = 7.66 x 3 6 2 3 x 3 V= 8 V = x3 2 3 x 12 4 SPHERICAL LUNE V= No. of VERTICES Formulas for VOLUME FRUSTUM OF A CONE Where: x = length of one edge SOLIDS WITH CURVED SURFACES CYLINDER is that portion of a spherical surface bounded by the halves of two great circles A( surface ) = πr 2θ (deg) 90° SPHERICAL ZONE is that portion of a spherical surface between two parallel planes. A spherical zone of one base has one bounding plane tangent to the sphere. A( zone ) = 2π r h V = Bh = KL SPHERICAL SEGMENT A(lateral) = PkL = 2 π r h is that portion of a sphere bounded by a zone and the planes of the zone’s bases. 2 A(surface) = A(lateral) + 2B Pk = perimeter of right section K = area of the right section B = base area L= slant height V = πh (3r − h) 3 πh (3a 2 + h 2 ) 6 πh V = (3a 2 + 3b 2 + h 2 ) 6 V = CONE 1 Bh 3 A(lateral ) = πrL V = SPHERICAL WEDGE is that portion of a sphere bounded by a lune and the planes of the half circles of the lune. V = PDF created with pdfFactory trial version www.pdffactory.com πr 3θ (deg) 270° SPHERICAL CONE PARABOLOID is a solid formed by the revolution of a circular sector about its one side (radius of the circle). a solid formed by rotating a parabolic segment about its axis of symmetry. 1 A( zone) r 3 A( surface) = A( zone) + A( lateralofcone ) V = SPHERICAL PYRAMID is that portion of a sphere bounded by a spherical polygon and the planes of its sides. 3 πr E V = 540° 1 V = πr 2 h 2 SIMILAR SOLIDS 3 3 V1 H R L = = = V2 h r l 2 2 A1 H R L = = = A2 h r l 2 E = [(n-2)180°] 3 A V1 = 1 A2 V2 E = Sum of the angles E = Spherical excess n = Number of sides of the given spherical polygon 2 3 ANALYTIC GEOMETRY 1 SOLIDS BY REVOLUTIONS RECTANGULAR COORDINATE SYSTEM TORUS (DOUGHNUT) a solid formed by rotating a circle about an axis not passing the circle. V = 2π2Rr2 A(surface) = 4 π2Rr ELLIPSOID 4 V = πabc 3 x = abscissa y = ordinate Distance between two points d = ( x2 − x1 ) 2 + ( y2 − y1 ) 2 Slope of a line m = tan θ = OBLATE SPHEROID a solid formed by rotating an ellipse about its minor axis. It is a special ellipsoid with c = a 4 V = πa 2b 3 y2 − y1 x2 − x1 Division of a line segment x= x1 r2 + x 2 r1 r1 + r2 y= y1r2 + y 2 r1 r1 + r2 y= y1 + y 2 2 PROLATE SPHEROID a solid formed by rotating an ellipse about its major axis. It is a special ellipsoid with c=b 4 V = πab 2 3 Location of a midpoint x= PDF created with pdfFactory trial version www.pdffactory.com x1 + x 2 2 STRAIGHT LINES Distance between two parallel lines General Equation d = Ax + By + C = 0 C1 − C2 A2 + B 2 Slope relations between parallel lines: m1 = m2 Point-slope form y – y1 = m(x – x1) Two-point form y − y1 = y2 − y1 ( x − x1 ) x2 − x1 Slope and y-intercept form Line 1 → Ax + By + C1 = 0 Line 2 → Ax + By + C2 = 0 Slope relations between perpendicular lines: m1m2 = –1 Line 1 → Ax + By + C1 = 0 Line 2 → Bx – Ay + C2 = 0 y = mx + b PLANE AREAS BY COORDINATES Intercept form A= x y + =1 a b Slope of the line, Ax + By + C = 0 m=− A B Angle between two lines m − m1 θ = tan −1 2 1 m m + 1 2 Note: Angle θ is measured in a counterclockwise direction. m2 is the slope of the terminal side while m1 is the slope of the initial side. Distance of point (x1,y1) from the line Ax + By + C = 0; d= 1 x1 , x2 , x3 ,....xn , x1 2 y1 , y2 , y3 ,.... yn , y1 Note: The points must be arranged in a counter clockwise order. LOCUS OF A MOVING POINT The curve traced by a moving point as it moves in a plane is called the locus of the point. SPACE COORDINATE SYSTEM Length of radius vector r: r = x2 + y2 + z2 Distance between two points P1(x1,y1,z1) and P2(x2,y2,z2) d = ( x2 − x1 ) 2 + ( y2 − y1 ) 2 + ( z 2 − z1 ) 2 Ax1 + By1 + C ± A2 + B 2 Note: The denominator is given the sign of B PDF created with pdfFactory trial version www.pdffactory.com Standard Equation: ANALYTIC GEOMETRY 2 (x – h)2 + (y – k)2 = r2 General Equation: CONIC SECTIONS a two-dimensional curve produced by slicing a plane through a three-dimensional right circular conical surface Ways of determining a Conic Section 1. 2. 3. 4. x2 + y2 + Dx + Ey + F = 0 Center at (h,k): h=− By Cutting Plane Eccentricity By Discrimination By Equation Radius of the circle: General Equation of a Conic Section: r 2 = h2 + k 2 − Ax2 + Cy2 + Dx + Ey + F = 0 ** Cutting plane Eccentricity Parallel to base e→0 Parallel to element e = 1.0 none e < 1.0 Parallel to axis e > 1.0 Discriminant Equation** Circle B2 - 4AC < 0, A = C A=C Parabola B2 - 4AC = 0 Ellipse B2 - 4AC < 0, A ≠ C Circle Parabola Ellipse Hyperbola Hyperbola B2 - 4AC > 0 D E ; k =− 2A 2A A≠C same sign Sign of A opp. of B A or C = 0 F 1 D2 + E 2 − 4F or r = A 2 PARABOLA a locus of a moving point which moves so that it’s always equidistant from a fixed point called focus and a fixed line called directrix. where: a = distance from focus to vertex = distance from directrix to vertex AXIS HORIZONTAL: Cy2 + Dx + Ey + F = 0 Coordinates of vertex (h,k): k=− CIRCLE A locus of a moving point which moves so that its distance from a fixed point called the center is constant. E 2C substitute k to solve for h Length of Latus Rectum: LR = PDF created with pdfFactory trial version www.pdffactory.com D C AXIS VERTICAL: ELLIPSE Ax2 + Dx + Ey + F = 0 Coordinates of vertex (h,k): h=− D 2A a locus of a moving point which moves so that the sum of its distances from two fixed points called the foci is constant and is equal to the length of its major axis. d = distance of the center to the directrix substitute h to solve for k Length of Latus Rectum: E LR = A STANDARD EQUATIONS: Major axis is horizontal: ( x − h)2 ( y − k ) 2 + =1 a2 b2 STANDARD EQUATIONS: Opening to the right: Major axis is vertical: 2 ( x − h) 2 ( y − k ) 2 + =1 2 2 b a (y – k) = 4a(x – h) Opening to the left: (y – k)2 = –4a(x – h) General Equation of an Ellipse: Ax2 + Cy2 + Dx + Ey + F = 0 Opening upward: (x – h) 2 = 4a(y – k) h=− Opening downward: (x – h) 2 = –4a(y – k) Latus Rectum (LR) a chord drawn to the axis of symmetry of the curve. LR= 4a Coordinates of the center: D E ;k = − 2A 2C If A > C, then: a2 = A; b2 = C If A < C, then: a2 = C; b2 = A KEY FORMULAS FOR ELLIPSE for a parabola Length of major axis: 2a Eccentricity (e) the ratio of the distance of the moving point from the focus (fixed point) to its distance from the directrix (fixed line). Length of minor axis: 2b Distance of focus to center: e=1 for a parabola c = a 2 − b2 PDF created with pdfFactory trial version www.pdffactory.com Length of latus rectum: 2 2b LR = a (y – k) = m(x – h) Transverse axis is horizontal: m=± Eccentricity: e= Equation of Asymptote: c a = a d b a Transverse axis is vertical: m=± HYPERBOLA a locus of a moving point which moves so that the difference of its distances from two fixed points called the foci is constant and is equal to length of its transverse axis. a b KEY FORMULAS FOR HYPERBOLA Length of transverse axis: 2a Length of conjugate axis: 2b Distance of focus to center: c = a 2 + b2 d = distance from center to directrix a = distance from center to vertex c = distance from center to focus Length of latus rectum: 2b 2 LR = a STANDARD EQUATIONS Transverse axis is horizontal ( x − h) (y − k) − =1 a2 b2 2 2 Eccentricity: e= c a = a d Transverse axis is vertical: ( y − k ) 2 ( x − h) 2 =1 − a2 b2 POLAR COORDINATES SYSTEM x = r cos θ GENERAL EQUATION Ax2 – Cy2 + Dx + Ey + F = 0 Coordinates of the center: h=− D E ; k =− 2A 2C If C is negative, then: a2 = C, b2 = A If A is negative, then: a2 = A, b2 = C PDF created with pdfFactory trial version www.pdffactory.com y = r sin θ r= x2 + y2 tan θ = y x SPHERICAL TRIGONOMETRY Napier’s Rules 1. The sine of any middle part is equal to the product of the cosines of the opposite parts. Co-op Important propositions 1. If two angles of a spherical triangle are equal, the sides opposite are equal; and conversely. 2. The sine of any middle part is equal to the product of the tangent of the adjacent parts. Tan-ad 2. If two angels of a spherical triangle are unequal, the sides opposite are unequal, and the greater side lies opposite the greater angle; and conversely. Important Rules: 3. The sum of two sides of a spherical triangle is greater than the third side. 2. When the hypotenuse of a right spherical triangle is less than 90°, the two legs are of the same quadrant and conversely. a+b>c 4. The sum of the sides of a spherical triangle is less than 360°. 1. In a right spherical triangle and oblique angle and the side opposite are of the same quadrant. 3. When the hypotenuse of a right spherical triangle is greater than 90°, one leg is of the first quadrant and the other of the second and conversely. 0° < a + b + c < 360° QUADRANTAL TRIANGLE 5. The sum of the angles of a spherical triangle is greater that 180° and less than 540°. 180° < A + B + C < 540° 6. The sum of any two angles of a spherical triangle is less than 180° plus the third angle. is a spherical triangle having a side equal to 90°. SOLUTION TO OBLIQUE TRIANGLES Law of Sines: sin a sin b sin c = = sin A sin B sin C A + B < 180° + C SOLUTION TO RIGHT TRIANGLES NAPIER CIRCLE Sometimes called Neper’s circle or Neper’s pentagon, is a mnemonic aid to easily find all relations between the angles and sides in a right spherical triangle. Law of Cosines for sides: cos a = cos b cos c + sin b sin c cos A cos b = cos a cos c + sin a sin c cos B cos c = cos a cos b + sin a sin b cos C Law of Cosines for angles: cos A = − cos B cos C + sin B sin C cos a cos B = − cos A cos C + sin A sin C cos b cos C = − cos A cos B + sin A sin B cos c PDF created with pdfFactory trial version www.pdffactory.com AREA OF SPHERICAL TRIANGLE π R2 E A= 180° R = radius of the sphere E = spherical excess in degrees, E = A + B + C – 180° TERRESTRIAL SPHERE Radius of the Earth = 3959 statute miles Prime meridian (Longitude = 0°) Equator (Latitude = 0°) Latitude = 0° to 90° Longitude = 0° to +180° (eastward) = 0° to –180° (westward) 1 min. on great circle arc = 1 nautical mile 1 nautical mile = 6080 feet = 1852 meters 1 statute mile = 5280 feet = 1760 yards 1 statute mile = 8 furlongs = 80 chains Derivatives dC =0 dx d du dv + (u + v ) = dx dx dx d dv du +v (uv ) = u dx dx dx dv du −u v d u dx = dx 2 dx v v du d n (u ) = nu n −1 dx dx du d u = dx dx 2 u du −c d c dx = 2 dx u u d u du (a ) = a u ln a dx dx d u du (e ) = e u dx dx du log a e d dx (ln a u ) = dx u du d (ln u ) = dx dx u d du (sin u ) = cos u dx dx d du (cos u ) = − sin u dx dx d du (tan u ) = sec 2 u dx dx d du (cot u ) = − csc 2 u dx dx d du (sec u ) = sec u tan u dx dx d du (csc u ) = − csc u cot u dx dx d du 1 (sin −1 u ) = 2 dx 1 − u dx PDF created with pdfFactory trial version www.pdffactory.com − 1 du d (cos −1 u ) = dx 1 − u 2 dx d 1 du (tan −1 u ) = dx 1 + u 2 dx d − 1 du (cot −1 u ) = dx 1 + u 2 dx d du 1 (sec −1 u ) = 2 dx u u − 1 dx d − 1 du (csc −1 u ) = dx u u 2 − 1 dx du d (sinh u ) = cosh u dx dx d du (cosh u ) = sinh u dx dx d du (tanh u ) = sec h 2 u dx dx d du (coth u ) = − csc h 2 u dx dx d du (sec hu ) = − sec hu tanh u dx dx d du (csc hu ) = − csc hu coth u dx dx d 1 du (sinh −1 u ) = dx u 2 + 1 dx d (cosh −1 u ) = dx 1 du u 2 − 1 dx d 1 du (tanh −1 u ) = dx 1 − u 2 dx d − 1 du (sinh −1 u ) = 2 dx u − 1 dx d − 1 du (sec h −1u ) = dx u 1 − u 2 dx d − 1 du (csc h −1u ) = dx u 1 + u 2 dx DIFFERENTIAL CALCULUS LIMITS Indeterminate Forms 0 , 0 ∞ , (0)(∞), ∞ - ∞, 0 0 , ∞ 0 , 1∞ ∞ L’Hospital’s Rule Lim x→a f ( x) f ' ( x) f "( x) = Lim = Lim ..... g ( x) x → a g ' ( x) x → a g" ( x) Shortcuts Input equation in the calculator TIP 1: if x → 1, substitute x = 0.999999 TIP 2: if x → ∞ , substitute x = 999999 TIP 3: if Trigonometric, convert to RADIANS then do tips 1 & 2 MAXIMA AND MINIMA Slope (pt.) Y’ MAX 0 Y” (-) dec Concavity down MIN 0 (+) inc up INFLECTION - No change - HIGHER DERIVATIVES nth derivative of xn dn (x n ) = n ! n dx nth derivative of xe n dn n X xe x n e ( ) = ( + ) dx n PDF created with pdfFactory trial version www.pdffactory.com TIME RATE the rate of change of the variable with respect to time + dx dt dx − dt = increasing rate = decreasing rate APPROXIMATION AND ERRORS If “dx” is the error in the measurement of a quantity x, then “dx/x” is called the RELATIVE ERROR. RADIUS OF CURVATURE 3 [1 + ( y ' ) 2 ] 2 R= y" INTEGRAL CALCULUS 1 ∫ du = u + C ∫ adu = au + C ∫ [ f (u) + g (u )]du = ∫ f (u )du + ∫ g (u)du u n+1 ∫ u du = n + 1 + C..............(n ≠ 1) du ∫ u = ln u + C au u a du = +C ∫ ln a n ∫ e du = e + C ∫ sin udu = − cos u + C ∫ cos udu = sin u + C ∫ sec udu = tan u + C ∫ csc udu = − cot u + C ∫ sec u tan udu = sec u + C ∫ csc u cot udu = − csc u + C u u 2 ∫ tan udu = ln sec u + C ∫ cot udu = ln sin u + C ∫ sec udu = ln sec u + tan u + C ∫ csc udu = ln csc u − cot u + C du ∫ = sin −1 u +C a a2 − u2 1 du −1 u ∫ a 2 + u 2 = a tan a + C 1 du −1 u ∫ u u 2 − a 2 = a sec a + C ∫ u = cos −1 1 − + C a 2au − u du 2 ∫ sinh udu = cosh u + C ∫ cosh udu = sinh u + C ∫ sec h udu = tanh u + C ∫ csc h udu = − coth u + C ∫ sec hu tanh udu = − sec hu + C ∫ csc hu coth udu = − csc hu + C ∫ tanh udu = ln cosh u + C ∫ coth udu = ln sinh u + C 2 2 ∫ ∫ du u +a du 2 u −a du 2 2 2 = sinh −1 u +C a = cosh −1 u +C a 1 a = − sinh −1 + C a u u +a du 1 −1 u ∫ a 2 − u 2 = a tanh a + C.............. u < a du 1 −1 u ∫ a 2 − u 2 = a coth a + C.............. u > a ∫u 2 2 ∫ udv = uv − ∫ vdu 2 PDF created with pdfFactory trial version www.pdffactory.com PLANE AREAS CENTROIDS Plane Areas bounded by a curve and the coordinate axes: Half a Parabola A= x2 ∫y ( curve) dx x1 y2 A = ∫ x( curve ) dy y1 Plane Areas bounded by a curve and the coordinate axes: x2 A = ∫ ( y( up ) − y( down ) )dx x1 3 x= b 8 2 y= h 5 Whole Parabola 2 y= h 5 Triangle y2 A = ∫ ( x( right ) − x(left ) )dy y1 Plane Areas bounded by polar curves: θ 1 2 2 A = ∫ r dθ 2 θ1 1 2 x = b= b 3 3 1 2 y= h= h 3 3 LENGTH OF ARC CENTROID OF PLANE AREAS (VARIGNON’S THEOREM) Using a Vertical Strip: x2 A • x = ∫ dA • x x2 ∫ S= x1 x1 x2 y A • y = ∫ dA • 2 x1 Using a Horizontal Strip: y2 x A • x = ∫ dA • 2 y1 y2 ∫ S= y1 S= z2 y2 A • y = ∫ dA • y y1 PDF created with pdfFactory trial version www.pdffactory.com ∫ z1 2 dy 1 + dx dx 2 dx 1 + dy dy 2 2 dx dy + dz dz dz INTEGRAL CALCULUS 2 CENTROIDS OF VOLUMES x2 V • x = ∫ dV • x x1 y2 TIP 1: Problems will usually be of this nature: • “Find the area bounded by” • “Find the area revolved around..” V • y = ∫ dV • y TIP 2: Integrate only when the shape is IRREGULAR, otherwise use the prescribed formulas WORK BY INTEGRATION Work = force × distance y1 VOLUME OF SOLIDS BY REVOLUTION x2 y2 x1 y1 W = ∫ Fdx = ∫ Fdy ; where F = k x Circular Disk Method x2 V = π ∫ R 2 dx Work done on spring Cylindrical Shell Method k = spring constant x1 = initial value of elongation x2 = final value of elongation 1 2 2 W = k ( x2 − x1 ) 2 x1 y2 V = 2π ∫ RL dy y1 Work done in pumping liquid out of the container at its top Circular Ring Method x2 V = π ∫ ( R − r )dx 2 Work = (density)(volume)(distance) 2 x1 Force = (density)(volume) = ρv PROPOSITIONS OF PAPPUS Specific Weight: First Proposition: If a plane arc is revolved about a γ = coplanar axis not crossing the arc, the area of the surface generated is equal to the product of the length of the arc and the circumference of the circle described by the centroid of the arc. Weight Volume A = S • 2π r γwater = 9.81 kN/m2 SI γwater = 45 lbf/ft2 cgs A = ∫ dS • 2π r Density: Second Proposition: If a plane area is revolved about a coplanar axis not crossing the area, the volume generated is equal to the product of the area and the circumference of the circle described by the centroid of the area. V = A • 2π r V = ∫ dA • 2π r ρ= mass Volume ρwater = 1000 kg/m3 SI ρwater = 62.4 lb/ft3 cgs ρsubs = (substance) (ρwater) 1 ton = 2000lb PDF created with pdfFactory trial version www.pdffactory.com MOMENT OF INERTIA Moment of Inertia about the x- axis: Ix = x2 ∫ y dA 2 x1 Moment of Inertia about the y- axis: Ellipse πab3 Ix = 4 πa 3b Iy = 4 y2 I y = ∫ x 2 dA FLUID PRESSURE Parallel Axis Theorem F = ∫ wh dA F = wh A = γ h A y1 The moment of inertia of an area with respect to any coplanar line equals the moment of inertia of the area with respect to the parallel centroidal line plus the area times the square of the distance between the lines. I x = Ixo = Ad 2 F = force exerted by the fluid on one side of the area h = distance of the c.g. to the surface of liquid w = specific weight of the liquid (γ) A = vertical plane area Moment of Inertia for Common Geometric Figures Specific Weight: Square γ = bh3 Ix = 3 I xo = 3 bh 12 γwater = 9.81 kN/m2 SI γwater = 45 lbf/ft2 cgs Triangle bh 3 Ix = 12 I xo = bh3 36 Circle I xo Weight Volume πr 4 = 4 Half-Circle πr 4 Ix = 8 Quarter-Circle πr 4 Ix = 16 PDF created with pdfFactory trial version www.pdffactory.com MECHANICS 1 CABLES PARABOLIC CABLES the load of the cable of distributed horizontally along the span of the cable. VECTORS Dot or Scalar product Uneven elevation of supports P • Q = P Q cosθ P • Q = Px Q x + Py Q y + Pz Q z 2 2 wx wx H= 1 = 2 2d1 2d 2 Cross or Vector product P × Q = P Q sin θ i j k P × Q = Px Qx Py Qy Pz Qz EQUILIBRIUM OF COPLANAR FORCE SYSTEM Conditions to attain Equilibrium: ∑F ∑F ∑M ( x − axis ) =0 ( y − axis ) =0 ( po int) =0 Friction Ff = μN tanφ = μ φ = angle of friction if no forces are applied except for the weight, φ=θ T1 = ( wx1 ) 2 + H 2 T2 = ( wx2 ) 2 + H 2 Even elevation of supports L > 10 d wL2 H= 8d 2 wL 2 T= +H 2 8d 2 32d 4 S = L+ − 3L 5 L3 L = span of cable d = sag of cable T = tension of cable at support H = tension at lowest point of cable w = load per unit length of span S = total length of cable PDF created with pdfFactory trial version www.pdffactory.com CATENARY MECHANICS 2 the load of the cable is distributed along the entire length of the cable. Uneven elevation of supports RECTILINEAR MOTION Constant Velocity T1 = wy1 T2 = wy 2 S = Vt H = wc Constant Acceleration: Horizontal Motion y1 = S1 + c 2 2 2 y2 = S 2 + c 2 2 2 S + y1 x1 = c ln 1 c S + y2 x 2 = c ln 2 c Span = x1 + x 2 1 2 at 2 V = V0 ± at S = V0 t ± V 2 = V0 ± 2aS 2 Total length of cable = S1 + S2 Even elevation of supports T = wy Constant Acceleration: Vertical Motion H = wc y = S +c 2 2 + (sign) = body is speeding up – (sign) = body is slowing down 2 ± H = V0t − 1 2 gt 2 S + y x = c ln c Span = 2 x V = V0 ± gt Total length of cable = 2S + (sign) = body is moving down – (sign) = body is moving up V 2 = V0 ± 2 gH 2 Values of g, general estimate exact PDF created with pdfFactory trial version www.pdffactory.com SI (m/s2) 9.81 9.8 9.806 English (ft/s2) 32.2 32 32.16 Variable Acceleration dS dt dV a= dt V= PROJECTILE MOTION ROTATION (PLANE MOTION) Relationships between linear & angular parameters: V = rω a = rα V = linear velocity ω = angular velocity (rad/s) a = linear acceleration α = angular acceleration (rad/s2) r = radius of the flywheel Linear Symbol Angular Symbol S V A t θ ω α t Distance Velocity Acceleration Time Constant Velocity x = (V0 cos θ )t ± y = (V0 sin θ )t − ± y = x tan θ − 1 2 gt 2 gx 2 2V0 cos 2 θ 2 Maximum Height and Horizontal Range V0 sin θ max ht 2g 2 y= 2 V sin 2θ x= 0 g 2 Maximum Horizontal Range Assume: Vo = fixed θ = variable 2 Rmax V = 0 ⇔ θ = 45° g θ = ωt Constant Acceleration 1 θ = ω 0 t ± αt 2 2 ω = ω 0 ± αt ω 2 = ω 0 ± 2αθ 2 + (sign) = body is speeding up – (sign) = body is slowing down D’ALEMBERT’S PRINCIPLE “Static conditions maybe produced in a body possessing acceleration by the addition of an imaginary force called reverse effective force (REF) whose magnitude is (W/g)(a) acting through the center of gravity of the body, and parallel but opposite in direction to the acceleration.” W REF = ma = a g PDF created with pdfFactory trial version www.pdffactory.com UNIFORM CIRCULAR MOTION motion of any body moving in a circle with a constant speed. mV 2 WV 2 Fc = = r gr V2 ac = r Fc = centrifugal force V = velocity m = mass W = weight r = radius of track ac = centripetal acceleration g = standard gravitational acceleration BANKING ON HI-WAY CURVES BOUYANCY A body submerged in fluid is subjected by an unbalanced force called buoyant force equal to the weight of the displaced fluid Fb = W Fb = γVd Fb = buoyant force W = weight of body or fluid γ = specific weight of fluid Vd = volume displaced of fluid or volume of submerged body Specific Weight: γ = Weight Volume γwater = 9.81 kN/m2 SI γwater = 45 lbf/ft2 cgs Ideal Banking: The road is frictionless V2 tan θ = gr Non-ideal Banking: With Friction on the road V2 tan(θ + φ ) = tan φ = µ ; gr V = velocity r = radius of track g = standard gravitational acceleration θ = angle of banking of the road φ = angle of friction μ = coefficient of friction Conical Pendulum T = W secθ 1 2π g h IMPULSE AND MOMENTUM Impulse = Change in Momentum F∆t = mV − mV0 F = force t = time of contact between the body and the force m = mass of the body V0 = initial velocity V = final velocity Impulse, I F V2 tan θ = = W gr f = ENGINEERING MECHANICS 3 Momentum, P frequency PDF created with pdfFactory trial version www.pdffactory.com I = F∆t P = mV Work, Energy and Power LAW OF CONSERVATION OF MOMENTUM “In every process where the velocity is changed, the momentum lost by one body or set of bodies is equal to the momentum gain by another body or set of bodies” Work W = F ⋅S Force Newton (N) Dyne Pound (lbf) Distance Meter Centimeter Foot Work Joule ft-lbf erg Potential Energy PE = mgh = Wh Momentum lost = Momentum gained Kinetic Energy KE linear = m1V1 + m2V2 = m1V1' + m2V2' m1 = mass of the first body m2 = mass of the second body V1 = velocity of mass 1 before the impact V2 = velocity of mass 2 before the impact V1’ = velocity of mass 1 after the impact V2’ = velocity of mass 2 after the impact KE rotational = 1 mV 2 2 1 2 Iω → V = rω 2 I = mass moment of inertia ω = angular velocity Coefficient of Restitution (e) e= Type of collision ELASTIC INELASTIC PERFECTLY INELASTIC Mass moment of inertia of rotational INERTIA for common geometric figures: V −V V1 − V2 ' 2 e 100% conserved Not 100% conserved Max Kinetic Energy Lost ' 1 Solid sphere: I = 2 2 mr 5 Kinetic Energy 0 < e >1 Thin-walled hollow sphere: e=0 e =1 Solid disk: I= 1 2 mr 2 I= m = mass of the body r = radius e= hr hd 2 2 mr 3 1 2 mr 2 1 2 2 Hollow Cylinder: I = m( router − rinner ) 2 Solid Cylinder: Special Cases I= e = cot θ tan β PDF created with pdfFactory trial version www.pdffactory.com Latent Heat is the heat needed by the body to change POWER its phase without changing its temperature. rate of using energy P= W = F ⋅V t 1 watt = 1 Newton-m/s 1 joule/sec = 107 ergs/sec 1 hp = 550 lb-ft per second = 33000 lb-ft per min = 746 watts LAW ON CONSERVATION OF ENERGY “Energy cannot be created nor destroyed, but it can be change from one form to another” Q = ±mL Q = heat needed to change phase m = mass L = latent heat (fusion/vaporization) (+) = heat is entering (substance melts) (–) = heat is leaving (substance freezes) Latent heat of Fusion – solid to liquid Latent heat of Vaporization – liquid to gas Values of Latent heat of Fusion and Vaporization, Kinetic Energy = Potential Energy WORK-ENERGY RELATIONSHIP The net work done on an object always produces a change in kinetic energy of the object. Work Done = ΔKE Positive Work – Negative Work = ΔKE Total Kinetic Energy = linear + rotation HEAT ENERGY AND CHANGE IN PHASE Sensible Heat is the heat needed to change the temperature of the body without changing its phase. Q = mcΔT Lf = 144 BTU/lb Lf = 334 kJ/kg Lf ice = 80 cal/gm Lv boil = 540 cal/gm Lf = 144 BTU/lb = 334 kJ/kg Lv = 970 BTU/lb = 2257 kJ/kg 1 calorie = 4.186 Joules 1 BTU = 252 calories = 778 ft–lbf LAW OF CONSERVATION OF HEAT ENERGY When two masses of different temperatures are combined together, the heat absorbed by the lower temperature mass is equal to the heat given up by the higher temperature mass. Q = sensible heat m = mass c = specific heat of the substance ΔT = change in temperature Specific heat values Cwater Cwater Cwater Cice Csteam = 1 BTU/lb–°F = 1 cal/gm–°C = 4.156 kJ/kg = 50% Cwater = 48% Cwater PDF created with pdfFactory trial version www.pdffactory.com Heat gained = Heat lost THERMAL EXPANSION For most substances, the physical size increase with an increase in temperature and decrease with a decrease in temperature. ΔL = LαΔT ΔL = change in length L = original length α = coefficient of linear expansion ΔT = change in temperature ΔV = VβΔT ΔV = change in volume V = original volume β = coefficient of volume expansion ΔT = change in temperature Note: In case β is not given; β = 3α THERMODYNAMICS In thermodynamics, there are four laws of very general validity. They can be applied to systems about which one knows nothing other than the balance of energy and matter transfer. ZEROTH LAW OF THERMODYNAMICS stating that thermodynamic equilibrium is an equivalence relation. If two thermodynamic systems are in thermal equilibrium with a third, they are also in thermal equilibrium with each other. FIRST LAW OF THERMODYNAMICS about the conservation of energy The increase in the energy of a closed system is equal to the amount of energy added to the system by heating, minus the amount lost in the form of work done by the system on its surroundings. SECOND LAW OF THERMODYNAMICS about entropy The total entropy of any isolated thermodynamic system tends to increase over time, approaching a maximum value. THIRD LAW OF THERMODYNAMICS, about absolute zero temperature As a system asymptotically approaches absolute zero of temperature all processes virtually cease and the entropy of the system asymptotically approaches a minimum value. This law is more clearly stated as: "the entropy of a perfectly crystalline body at absolute zero temperature is zero." PDF created with pdfFactory trial version www.pdffactory.com STRENGTH OF MATERIALS SI N/m2 = Pa SIMPLE STRESS Stress = Units of σ Force Area mks/cgs English Kg/cm2 lbf /m2 = psi kN/m2 = kPa 103 psi = ksi Axial Stress MN/m2 = MPa 103 lbf = kips the stress developed under the action of the force acting axially (or passing the centroid) of the resisting area. GN/m2 = Gpa σ axial P = axial A N/mm2 = MPa Standard Temperature and Pressure (STP) Paxial ┴ Area σaxial = axial/tensile/compressive stress P = applied force/load at centroid of x’sectional area A = resisting area (perpendicular area) 101.325 kPa Shearing stress the stress developed when the force is applied parallel to the resisting area. = = = = = = = 14.7 psi 1.032 kgf/cm2 780 torr 1.013 bar 1 atm 780 mmHg 29.92 in P σs = A Thin-walled Pressure Vessels Pappliedl ║ Area σT = σs = shearing stress P = applied force or load A = resisting area (sheared area) A. Tangential stress B. Longitudinal stress (also for Spherical) Bearing stress the stress developed in the area of contact (projected area) between two bodies. σb = ρ r ρD = t 2t σL = ρ r ρD = 2t 4t P P = A dt P ┴ Abaering σb = bearing stress P = applied force or load A = projected area (contact area) d,t = width and height of contact, respectively σT = tangential/circumferential/hoop stress σL = longitudinal/axial stress, used in spheres r = outside radius D = outside diameter ρ = pressure inside the tank t = thickness of the wall F = bursting force PDF created with pdfFactory trial version www.pdffactory.com SIMPLE STRAIN / ELONGATION Types of elastic deformation: Strain – ratio of elongation to original length a. Due to axial load HOOKE’S LAW ON AXIAL DEFORMATION “Stress is proportional to strain” σαε σ = Yε Young ' s Modulus of Elasticity σ = Eε Modulus of Elasticity σ s =E sε s Modulus in Shear σ V =EV εV 1 Ev Bulk Modulus of Elasticity compressibility δ = ε= ε = strain δ = elongation L = original length δ L Elastic Limit – the range beyond which the material WILL NOT RETURN TO ITS ORIGINAL SHAPE when unloaded but will retain a permanent deformation Yield Point – at his point there is an appreciable elongation or yielding of the material without any corresponding increase in load; ductile materials and continuous deformation Ultimate Strength – it is more commonly called ULTIMATE STRESS; it’s the hishes ordinate in the curve Rupture Strength/Fracture Point – the stress at failure PL AE δ = elongation P = applied force or load A = area L = original length E = modulus of elasticity σ = stress ε = strain b. Due to its own mass ρgL2 mgL δ = = 2E 2 AE δ = elongation ρ = density or unit mass of the body g = gravitational acceleration L = original length E = modulus of elasticity or Young’s modulus m = mass of the body c. Due to changes in temperature δ = Lα (T f − Ti ) δ = elongation α = coefficient of linear expansion of the body L = original length Tf = final temperature Ti = initial temperature PDF created with pdfFactory trial version www.pdffactory.com d. Biaxial and Triaxial Deformation µ=− εy εx =− εz εx Power delivered by a rotating shaft: P = Tω Prpm = 2πTN rps TORSIONAL SHEARING STRESS 2πTN 60 2πTN Php = 550 2πTN Php = 3300 Torsion – refers to twisting of solid or hollow rotating shaft. T = torque N = revolutions/time Solid shaft HELICAL SPRINGS μ = Poisson’s ratio μ = 0.25 to 0.3 for steel = 0.33 for most metals = 0.20 for concrete μmin = 0 μmax = 0.5 Prpm = ft − lb sec ft − lb min 16T πd 3 τ = 16 PR d 1 + πd 3 4 R 16TD π (D 4 − d 4 ) τ = 16 PR 4m − 1 0.615 + m πd 3 4m − 4 τ= Hollow shaft τ = rpm τ = torsional shearing stress T = torque exerted by the shaft D = outer diameter d = inner diameter where, Maximum twisting angle of the shaft’s fiber: elongation, θ= TL JG θ = angular deformation (radians) T = torque L = length of the shaft G = modulus of rigidity J = polar moment of inertia of the cross πd 4 J= 32 m= Dmean Rmean = d r 64 PR 3 n δ = Gd 4 τ = shearing stress δ = elongation R = mean radius d = diameter of the spring wire n = number of turns G = modulus of rigidity → Solid shaft π (D 4 − d 4 ) J= → Hollow shaft 32 Gsteel = 83 GPa; Esteel = 200 GPa PDF created with pdfFactory trial version www.pdffactory.com Mode of Interest Annually Semi-Annually Quarterly Semi-quarterly Monthly Semi-monthly Bimonthly Daily ENGINEERING ECONOMICS 1 SIMPLE INTEREST I = Pin F = P (1 + in) P = principal amount F = future amount I = total interest earned i = rate of interest n = number of interest periods m 1 2 4 8 12 24 6 360 Shortcut on Effective Rate Ordinary Simple Interest n= days 360 n= months 12 Exact Simple Interest days → ordinary year 365 days n= → leap year 366 n= ANNUITY Note: interest must be effective rate COMPOUND INTEREST F = P(1 + i ) n Nominal Rate of Interest i= NR ⇔ n = mN m Effective Rate of Interest ER = (1 + i ) − 1 m Ordinary Annuity A [(1 + i ) n − 1] F= i A [(1 + i ) n − 1] P= (1 + i ) n i m NR ER = 1 + −1 m ER ≥ NR ; equal if Annual i = rate of interest per period NR = nominal rate of interest m = number of interest periods per year n = total number of interest periods N = number pf years ER = effective rate of interest A = uniform periodic amount or annuity Perpetuity or Perpetual Annuity P= PDF created with pdfFactory trial version www.pdffactory.com A i LINEAR / UNIFORM GRADIENT SERIES ENGINEERING ECONOMICS 2 DEPRECIATION P = PA + PG Straight Line Method (SLM) d= G (1 + i) n − 1 n PG = − i i(1 + i) n (1 + i) n FG = Dm = md G (1 + i ) n − 1 − n i i Cm = C0 – Dm d = annual depreciation C0 = first cost Cm = book value Cn = salvage or scrap value n = life of the property Dm = total depreciation after m-years m = mth year 1 n AG = G − n i (1 + i) − 1 Perpetual Gradient PG = C0 − Cn n G i2 Sinking Fund Method (SFM) UNIFORM GEOMETRIC GRADIENT d= (C 0 − C n )i (1 + i ) n − 1 d [(1 + i ) m − 1] Dm = i Cm = C0 – Dm (1 + q) n (1 + i ) − n − 1 P = C q−i if q ≠ i (1 + q) n − (1 + i ) n F = C q−i if q ≠ i Cn P= 1+ q q= Cn(1 + i) n P= 1+ q i = standard rate of interest Sum of the Years Digit (SYD) Method 2(n − m + 1) d m = (C 0 − C n ) n(n + 1) if q = i sec ond −1 first C = initial cash flow of the geometric gradient series which occurs one period after the present q = fixed percentage or rate of increase (2n − m + 1)m Dm = (C 0 − C n ) n( n + 1) SYD = n(n + 1) 2 Cm = C0 – Dm SYD = sum of the years digit dm = depreciation at year m PDF created with pdfFactory trial version www.pdffactory.com Declining Balance Method (DBM) k =1− n BONDS P = Panuity = Pcpd int erest Cn Co Zr[(1 + i ) n − 1] C P= + n (1 + i ) i (1 + i) n Matheson Formula C k =1− m m Co C m = C 0 (1 − k ) m d m = kC 0 (1 − k ) m −1 Dm = C0 – Cm k = constant rate of depreciation P = present value of the bond Z = par value or face value of the bond r = rate of interest on the bond per period Zr = periodic dividend i = standard interest rate n = number of years before redemption C = redemption price of bond BREAK-EVEN ANALYSIS Total income = Total expenses CAPITALIZED AND ANNUAL COSTS CC = C 0 + P CC = Capitalized Cost C0 = first cost P = cost of perpetual maintenance (A/i) AC = d + C 0 (i ) + OMC AC = Annual Cost d = Annual depreciation cost i = interest rate OMC = Annual operating & maintenance cost PDF created with pdfFactory trial version www.pdffactory.com