Appreciation For Pre Calculus

APPRECIATION
>Pre Calculus is a course in mathematics that prepares students like us for calculus, It includes algebra,trigonometry and geometry Pre Calculus is an introduction or a "gateway" to a deeper and challenging study of mathematics known as Calculus.

APPLICATIONS
Calculus can be applied to determine the cardiac output (the amount of blood pumped by heart every minute)

>When you’re working in the kitchen, there’s an immense amount of calculus and precalculus going on in addition to basic cooking arithmetic. The currents when you boil pasta, and other forms of heat transformation (how to keep something from getting cold?) are closely dependent on differential equations.

 

Impact Testing For Cars

The distance and the time is measured in this situations, this saves peoples lives every year because of prevention of accidents

>Calculating The Rate Of an Outbreak

Pre Calculus can be used in the medical field to calculate the rate at which disease would spread. The equation that we would use would be logistical equation

> Overall Pre Calculus Is essential to our daily lives especially to us students who were aspiring to be an engineer or a doctor, or a chemist.

>To me It is a great opportunity to have this subject to study in STEM strand. Even this subject gives me temporary deppressions, breakdowns I am very thankful that our teacher introduced Pre Calculus to us  because this can help us to prepare for an even harder, challenging courses that we might face in the future. Thank You Sir Pre Calculus and God Bless... :)

Examples of Parabolas

5 SOLVED PROBLEMS OF PARABOLA

EXAMPLE#1

Find the focus and the directrix of

The following parbola

y² = 16x

x = 16y²

4p = 16

4        4

p = 4

Focus = (h+4,0) Directrix: y=k – 4

Focus = (4,0) Directrix: x = -4

GRAPH:

EXAMPLE#2

Find the directrix,focus, and vertex of

The parabola

y = ½ x²

(x – h)² = 4p(y – k) Standard Form

1/2 x² = y

x² = 2y

2 = 4p

4    4

p = ½

(x – 0)² = 4(1/2)(y – 0)²

Vertex (0,0) Focus (0,1/2) Directrix: y = -1/2

GRAPH:

EXAMPLE#3

Determine the vertex,focus, and the

Directrix of the parbola from this

Equation and graph

(y – 1)² = 8(x + 5)

Vertex (h,k) = (-5,1)

Focus = 4p = 8

              4      4

p = 2 Focus: (-5+2,1)

Focus: (-3,1)

Directrix: x = - 5 - 2

Directrix: x = -7

GRAPH:

EXAMPLE#4

Determine the diractrix , focus and

The vertex of the parabola and graph

(x – 3)² = 3(y + 1)

Vertex (h,k) = (3,-1)

4p = 3

4      4

Focus (3,-1 + ¾) Focus (3, - ¼)

Directrix: y = 1 – ¾  Directrix: y = -7/4

GRAPH:

EXAMPLE#5

Find the Standard Form of a parabola

from the given directrix and vertex and

graph

Vertex (0,4) Directrix: y = 2

SOLUTION

P = 2

Focus = (0,4+2) Focus (0,6)

(x-h)² = 4p(y-4)

(x-0)² = 4(2)(y – 4)

x² = 8(y – 4)

x² = 8y – 32

y = 1/8 (x² + 32) S.F

GRAPH:

And that's all :)

Examples Of Circles

5 SOLVED PROBLEMS OF CIRCLE

EXAMPLE# 1

Find the center and radius of this circle:

x² + y² – 8x + 2y – 19 = 0

SOLUTION:

x² – 8x + 16 + y² + 2y + 1 = 19+16+1

(x – 4)² + (y + 1)² = ⎷36

(x – 4)² + (y + 1)² = 6²

Center: (4,-1) Radius: r = 6
GRAPH:

EXAMPLE#2

What is the center and radius of the

circle indicated by this equation?

(x – 2)² + (y)² = 36

SOLUTION:

(x – 2)² + (y)² = ⎷36

(x – 2)² + (y)² = 6²

Center: (2,0) Radius: r = 6

GRAPH:

EXAMPLE#3

What is the equation of the circle

with a center at (4,-5) and a point

in the circle at (4,-2).What is the

radius?

SOLUTION:

(x-h)² + (y-k)² = r² Standard Formula

(x-4)² + (y-(-5))² = r² 

(4-4)² + (-2+5)² = r²

(0)² + (3)² = r²

9 = r²

r = 3²

^GRAPH:

Center: (4,-5) Radius: r = 3

EXAMPLE#4

Determine the center and the radius

Of the circle from this equation and graph

x² + y² – 8x + 6y + 9 = 0

SOLUTION:

x² – 8x + 16 + y² + 6y + 9 = -9 + 9 + 16

(x-4)² + (y+3)² = 16

(x-4)² + (y+3)² = 16

(x-4)² + (y+3)² = 4²

Center: (4,-3) Radius: r = 4

GRAPH:

EXAMPLE#5

Determine the center and the radius

Of the circle from this equation and

graph

x² + y² – 6x + 6y – 18 = 0

x² – 6x + 9 + y² + 6y + 9 = 18 + 9 + 9

(x-3)² + (y+3)² = 36

(x-3)² + (y+3)² = 36

(x-3)² + (y+3)² = 6²

Center: (3,-3) Radius: r = 6

GRAPH:

1st Lesson Conic Sections

1st Lesson: Conic Sections

The conic sections are the nondegenerate curves generated by the intersections of a plane with one or two nappes of a cone. For a plane perpendicular to the axis of the cone, a circle is produced. For a plane that is not perpendicular to the axis and that intersects only a single nappe, the curve produced is either an ellipse or a parabola (Hilbert and Cohn-Vossen 1999, p. 8). The curve produced by a plane intersecting both nappes is a hyperbola (Hilbert and Cohn-Vossen 1999, pp. 8-9

GENERAL EQUATION
Ax²­+Bxy+Cy²+Dx+Ey+F=0


THE DISCRIMINANT FORMULA FOR CONIC SECTION

 b² – 4ac > 0 HYPERBOLA
b²  - 4ac = 0 PARABOLA
b²  - 4ac <  0 CIRCLE

         where A = C B = 0
b²  - 4ac <  0 ELLIPSE
         where A≠c B=0

EXAMPLES:
1. G.E
 2y² + 20x + 2y – 1 = 0
 A= 0  B=  0  C= 2
b² – 4ac = (0)⁰ – 4 (0)(2)
                = 0 – 0
                = 0 PARABOLA

S.F : (h,k) (y-k)² = 4a(x-h)

2y² + 20x + 12y – 1 = 0

2y² + 12y       20x – 1 = 0

2(y² + 6y + 9) = -20x + 1

 2(y+3)²  = -20x + 1 + 9

    2

(y+3)² = -10x + 10

S.F = (y+3)² = -10x + 10

h = 5 k = 3

 2. G.E

6x² – 5y + 36x + 20y – 16 = 0

A= 6  B= 0  C= -5

b² – 4ac = (0)² – 4 (6)(-5)

= 120 < 0  HYPERBOLA

S.F

6x² + 36x ____ -5y + 20y – 16 = 0

6(x²+6x+9) – 5(y² – 4y + 4)= 16 + 9 + 4

6(x+3)² – 5(y-2)² = 29

              30

(x +3)² – (y-2)²

     5            6

   10          10

(h = 3) (k= -2)