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Partial Fraction Rules

1. One constant at a time,

2. Disregard the other constants and their divisors if it’s not the one that you are solving.

3. Get the factor/s or the “X” value of the denominator of the constant that you are solving.

Note: Usually the factors or the value of “X” is an imaginary number, especially if the

denominator of the constants that you are solving is a quadratic equation.

USE MODE 5-3 or the usual equate to zero then Find “X” teknik.

4. Free the constant of its denominator by multiplying its denominator to both sides of the equation.

5. Use Mode 2 then solve for the value of the left side of the equations by substituting the value of

“X” from Step 3. (pwede din naming hindi mode 2 pag walang complex or imaginary number)

Note: The maximum variable that the complex mode (Mode 2) can solve is only up to .

(Testingin nyo pa)

So you make = ; or =

- If the constant that you are solving also has the variable “X”. Substitute the value of “X” too,

from step 3.

e.g.

, value of X here = √ and √ you can use either of the two.

Using x = √ , So = (D√ + E )

6. IMPORTANT!! (this is the usual thing that people forget when solving partial fractions,

this is the most important thing to remember!!)

- Complex is for Complex, Real numbers are for Real Numbers.

Sample Problem.

May 2004 Problem: What is the value of E in the following Equation.

Given :

=

Solution:

Solving for E.

Find x from denominator of (Dx+E) which is , using Mode5-3,

[1 0 3] = √ and √ (you can use either of them) I will use .

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First term of the polynomial from the left side equation has an exponent of 4, since you have an imaginary

number as a root, you need to use complex mode (Mode 2). (See Note from Step 5).

Solving ( so disregard.

Focus on this Equation for the mean time.

=

Multiply on both sides of the equation gives.

=

Solve for the value of the left side equation with √ (MODE 2) from .

This is exactly what you enter on your calculator.

; calc. X?= enter √

= -1-2√

Also substitute √ to the right side of the equation.

Substituting gives.

= √

The left side and the right side equations now gives.

-1-2√ = √

REMEMBER THE MOST IMPORTANT RULE IN PARTIAL

FRACTIONS. Complex for Complex, Real numbers for Real Numbers

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Solving simultaneously:

Complex is for Complex:

-2√ = √ ;

-2 = √

√

D= -2

Real numbers for Real Numbers:

-1 =

E = -1

Additional Questions for practice. Solve for the values of A, B, and C.

Answers.

A =2

B =0

C = 3

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Geotech Caltek:

EQUIVALENT HYDRAULIC CONDUCTIVITY IN STRATIFIED SOIL.

=∑

(Horizontal)

Caltech.

Mode 3-1.

X f

X = k

F = depth

x =

(shift 1-4-2)

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=

∑

(Vertical)

Caltech.

Mode 3-1.

X f

X = (Inverse of k)

F = depth

x = = (inverse of x bar)

(shift 1-4-2)

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Mohr Circle Technique.

Alam nyo naman na siguro yung sign convention.

Sa stress

compression = (+)

Tension = (-)

Sa Shear

Counter Clock Wise = (+)

Clock Wise = (-)

Technique. (Go to Complex Mode) Mode 2.

) store to A

) store to B

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Store to C

| | = (Radius) Store to D

arg(A-C) = Store to E

arg( ) is Shift 2-1 of (complex Mode)

max stress = C + D

min stress = C – D

normal ( and shear stress ( on plane

= C + D ( E + 2 )

Anglev is Shift A

Practice Problem on MegaReview Module. Geotech.

Stresses in Soil.

Problem #1.

Mode 2.

(250+0i) A

(90+0i) B

170 C

| | 80 D

arg(A-C) = 0

max stress = C + D = 250

min stress = C – D = 90

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normal ( and shear stress ( on plane

= C + D ( 60 ) 60 is from 2

= 210 +69.2820323i

69.2820323

Problem #2.

Mode 2.

(300 + 40i) A

(120 - 40i) B

210 C

| | 98.49 D

arg(A-C) = 23.96 E

max stress = C + D = 308.49

min stress = C – D = 111.51

normal ( and shear stress ( on plane

= C + D ( 40 ) 40 is from 2

= 253.2324955 +88.4926626i

88.4926626