Transcript
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
