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Saturday, January 19, 2013

First Order Linear Equation

Category: Differential Equations, Integral Calculus, Algebra

"Published in Newark, California, USA"

Find the general solution for the given equation



Solution:

Consider the given equation



Check if the given equation is exact or not exact as follows

Let 



Let



Since 

The given equation is not an exact equation. In this case, we need to find the integrating factor and then multiply it to the both sides of the equation. Let's consider again the given equation



Arrange the above equation into its standard form as follows





Divide both sides of the equation by dx



Divide both sides of the equation by (x + 1)2



As you notice that the above equation is now already in standard form



where
and

The integrating factor is







Therefore, the general solution for the above equation is












Friday, January 18, 2013

Area Derivation - Triangle, Three Vertices

Category: Analytic Geometry, Plane Geometry, Algebra

"Published in Newark, California, USA"

Given a triangle with vertices as shown below


Photo by Math Principles in Everyday Life

Prove that the area of a triangle is



Solution:

The first step is to draw vertical lines from the vertices to the x-axis as shown below


Photo by Math Principles in Everyday Life


Area of Triangle ABC = Area of Trapezoid ACDF - Area of Trapezoid ABEF - Area of Trapezoid BCDE


Label further the figure, we have


Photo by Math Principles in Everyday Life

Consider Trapezoid ACDF







Consider Trapezoid ABEF







Consider Trapezoid BCDE







Therefore










Using a distance of two points formula











The above equation becomes












You notice that the above equation looks like a matrix or determinant because of the sequence of x and y. The three positive terms are the principal diagonals (product from top left to bottom right) while the three negative terms are the secondary diagonals (product from bottom left to top right). The above equation can be written as



Therefore




Thursday, January 17, 2013

Proving - Parallelogram, Triangles

Category: Plane Geometry

"Published in Newark, California, USA"

Given: Parallelogram CDEF; S and T are midpoints of EF and ED. 

Prove: SR FD


Photo by Math Principles in Everyday Life

Solution:

Consider the given figure


Photo by Math Principles in Everyday Life

Proof:

1. Statement: Parallelogram CDEF

    Reason: Given item.

2. Statement: S and T are the midpoints of EF and ED.

    Reason: Given item.

3. Statement: FS SE 

    Reason: Point S bisects the line segment EF into two equal parts. 

4. Statement: ET ≅ TD

    Reason: Point T bisects the line segment ED into two equal parts.

5. Statement: FE ║ CD

    Reason: The opposite sides of a parallelogram are parallel.

6. Statement: SET ≅ TDR

    Reason: If a transversal line (ED) passed the two parallel lines (EF and CD), then the alternating interior angles are congruent. 

7. Statement: ∠ STE ≅ ∠ DTR

    Reason: Vertical angles are congruent.

8. Statement: ∆SET ≅ DTR

    Reason: Angle Side Angle (ASA) Postulate.

9. Statement: DR SE

    Reason: Since ∆SET ≅ DTR, then all sides of a triangle are congruent to all sides of other triangle.

10. Statement: DR ≅ SE FS

      Reason: Transitive property of congruence. 

11. Statement: FS ║ DR

      Reason: Since Point R is colinear with CD and CD is parallel to EF, then it follows that FS is parallel to DR since Point S is colinear with EF.

12. Statement: FDRS is a parallelogram.

      Reason: Since FS and DR are parallel and congruent, then it follows that the figure formed by the points FDRS is a parallelogram.

13. Statement: FD SR

      Reason: The two opposite sides of a parallelogram are congruent and parallel.