Friday, January 25, 2013

Inverse Cosine Graph

Inverse cosine  is one of the essential  inverse trigonometric function . Its principal values and range is
cos -1 : {(y,x)l y = cos x , x `in [ 0, pi]` , y `in` [-1,1]}


In this we deals with determining inverse tangent graphs.

Function        Domain                      Principal Value (Range)
y =cos -1 x             R                            [ 0, `pi`]

The graph of cos-1 x is as follows :

. inverse cosine graph1

Inverse Cosine Graph : Examples

Example  1 : Draw the inverse cosine graph for the function
                                        y = cos-1 ( x + 4 ) 
Solution : As it is given   y= cos -1 ( x + 4 ) so, there is a horizontal shift in the graph of   y = cos-1 x by 4 units leftwards 

Graph as shown :

                  inverse cosine graph2

Example  2  : Draw the inverse cosine graph for the function
                                       y = cos-1 ( x ) - 3 
Solution : As it is given   y = cos-1 ( x ) - 3

It can also be wirtten as  y + 3  = cos-1 ( x ) so , there is a vertical shift in the graph of   y =cos -1 x by 3 unit downwards

Graph as shown :

                    inverse cosine graph3

Example  3  : Draw the inverse cosine graph for the function
                                        y = cos-1 ( x + 2 ) and   y = cos-1 ( x - 1 ) on the same graph 
Solution : As it is given   y = cos-1 ( x + 2 ) so, there is a horizontal shift in the graph of   y = cos-1 x by 2 units leftwards and

               for  y = cos -1 ( x - 1 ) there is a horizontal shift in the graph of   y = cos-1 x by 1 unit rightwards
Graph as shown :

             inverse cosine graph4

Inverse Cosine Graph : Practice Problems

Problem 1  :  Draw the inverse cosine graph for the function
                                        y = cos-1 ( x + 5 )
Problem  2 :  Draw the inverse cosine graph for the function               
                                             y = cos-1 ( x ) + 6
Problem  3  :  Draw the inverse cosine graph for the function
                                        y = cos-1 ( x + 1) and   y = cos-1 ( x - 7 ) on the same graph

Thursday, January 24, 2013

Number Line Estimation

In mathematics, a line with points marked on it is termed as number line. Every point represents a number in number line. This number line is mainly used for representing the numbers. From online, we have a clear description of number line estimation. This article gives the explanation of number line estimation and some example problems using number line.

Explanation to Number Line Estimation:

The facts of number line is as follows.
  • A straight horizontal line with points that are evenly spaced.
  • A number o is at center and positive numbers are at right side of 0 while the negative numbers are at left side of 0.
  • The number line makes easy the arithmetic operations addition and subtraction.
A number line with points is as follows.

Number line

Let us see some example problems using number line.

Example Problems to Number Line Estimation:

Example: 1
Estimate a number 5 between the numbers 0 to 8.

Solution:
The estimation of a number 5 is as follows.

Estimation of 5 on number line

Example: 2
Estimate a negative number - 5 between -9 to 0.

Solution:
The estimation of a negative number -5 is as follows.
Estimation of -5 on number line

Example: 3
Add the numbers 1 and 6 using number line.

Solution:
Given: 1 + 6

Step 1:
Mark 1 on number line.
Mark 1 on number line
Step 2:
Start count and move from 1 at right side.
Addition on number line
Step 3:
Stop count when it reaches 6 and mark the resultant value.

Addition on number line

Answer: 7

Example: 4
Subtract 2 from 4 using number line.

Solution:
Given: 4 - 2

Step 1:
Mark a number 4 on number line.

Subtraction on number line

Step 2:
Start count and move from 4 at left side.

Subtraction on number line

Step 3:
Stop count when it reaches 2 and mark the resultant value.

Subtraction on number line

Answer: 2

Practice Problems to Number Line Estimation:

Problem: 1
Add the numbers 2 and 6 using number line.

Answer: 8

Problem: 2
Subtract 3 from 4 using number line.

Answer: 1

Wednesday, January 23, 2013

Square Root Function Problems


In general, square root of an integer A is a number B such that B2 = A, and we also deliver it by another ways, a number B whose square is A. Every non-unenthusiastic real number A has a only one of its kind of non-unenthusiastic square root, is said to be as the principal square root root, defined by a square roots symbol as sqrt (x). For positive (+) A, the principal square root will also rewrite in exponent notation, as like A1/2. For e.g. Principal square root of 16 is 4, obtained by sqrt (16) = 4, because 42 = 4 × 4 = 16 and 4 is non-negative.

Different Functions of Square Root:

Square root in math having different properties they are described below,

General expression with exponent and radical: `(^nsqrt (a) ^m) =(^nsqrt (a)) ^m =(a1/n) ^m = a^m/n`

Multiplication property for radical expression: `(^nsqrt (ab))= (^nsqrt (a)) (^nsqrt (b))`

Division property for radical expression: `(^nsqrt (a/b)) = (^nsqrt (a)) / (^nsqrt (b))`

Square Root Function Problem:

Example for square root in math: Add: `(9 xx sqrt (5))-(6 xx sqrt (5)) + (8 xx sqrt (5))`

Solution: Unite like terms by adding together the numerical coefficients.

`(9 - 6 + 8) xx sqrt (5)`

`(17 - 6) xx sqrt (5)`

After adding together, we get the answer like,

`11 xx sqrt (5)`

Example for square root function:   Simplify `sqrt (81) + sqrt (9)`

Solution: Take the given question and split the terms like,

`sqrt (9 xx 9) + sqrt (3 * 3)`

Rewrite the square root by using product of square root theorem.

`sqrt (9) xx sqrt (9) + sqrt (3) xx sqrt (3)`

To simplify even further we use the definition of square root,

`= 9 + 3`

Simplify by adding like terms to get the answer

`= (9 + 3)`

`= 12`

Thus, `sqrt (81) + sqrt (9) = 12.`

Example for square root function: Solve `sqrt (2) xx sqrt (196) = sqrt (392)`

` = sqrt (2 xx 196)`

`= sqrt (2 xx 14^2)`

` = ^14 sqrt (2)`

Practice problem for square root function:

Problem for square root function: `(sqrt (121))`

Answer: `11`

Problem for square root function: `(sqrt (169))`

Answer: `13`

Problem for square root function: `(sqrt (2025))`

Answer: `45`

Problem for square root function: `sqrt (2) + sqrt (4)`

Answer: `2 sqrt (2)`

Monday, January 21, 2013

Trig Ratio Table

Special angles comprise trigonometric values that may be considered exactly. The special angle values can be calculated using trigonometric ratios. Specifically, the angles of 0, 30, 45, 60, 90 degrees are the special ones. Right angles include the measure as 90 degree. A right triangle consists of a Right Angle as one of its 3 angles. The right triangles consist of two special triangles which are 30-60-90 right triangle and 45-45-90 right triangle. Let us see about the trig ratio table.

Trig Ratio Table


Followings are the six trigonometric ratios

Sin θ = `(opposite)/("hypotenuse")`

Cos θ = `(adjacent)/("hypotenuse")`

Tan θ = `(opposite)/(adjacent)`

Cosec θ = `("hypotenuse")/(opposite)`

Sec θ = `("hypotenuse")/(adjacent)`

Cot θ = `(adjacent)/(opposite)`

Trig ratio table

In the  diagram represents the trig ratio table.


Example

Evaluate 46 sin 30˚ + 16 cos 60˚ - 8 tan 45˚

Solution

46 sin 30˚ + 16 cos 60˚ - 8 tan 45˚

46 x `1/2`   + 16 x `1/2`  - 8

23+ 8 -8

31 - 8 = 23

46 sin 30˚ + 16 cos 60˚ - 8 tan 45˚ = 23

Examples for Trig Ratio Table

Example 1 for trig ratio table

Find the adjacent side of the right angle triangle if angle θ is 60°, and the hypotenuse is 24.

Solution:

Cos θ =`(adjacent)/("hypotenuse")`

Given

Given θ is 60 degree

Hypotenuse side = 24

Assume adjacent side = x

Cos 600 =  `x/24`

x=   `24 xx cos 60^0`

x =  `24xx(1/2)`

x = 24 x 0.5        {Value of cos 60 degree`1/2` }

x = 12

Therefore, the adjacent side of the right angle triangle = 12

Example 2 for trig ratio table

Evaluate 18 sin 30˚ + 16 cos 60˚ - 8 tan 45˚

Solution

18 sin 30˚ + 16 cos 60˚ - 8 tan 45˚

18 x `1/2`   + 16 x `1/2`  - 8

9+ 8 -8

17 - 8 = 9

18 sin 30˚ + 16 cos 60˚ - 8 tan 45˚ = 9

Example 3 for trig ratio table

Evaluate 28 sin 30˚ + 16 cos 60˚ - 8 tan 45˚

Solution

28 sin 30˚ + 16 cos 60˚ - 8 tan 45˚

28 x `1/2`   + 16 x `1/2`  - 8

14+ 8 -8

22 - 8 = 14

28 sin 30˚ + 16 cos 60˚ - 8 tan 45˚ = 14

Example 4 for trig ratio table

Evaluate 12 cosec 30˚ + 16 sec 60˚ - 8 cot 45˚

Solution

12 cosec 30˚ + 16 sec 60˚ - 8 cot 45˚

12 x 2 + 16 x 2- 8

24+ 32 -8

56 - 8 = 48

12 cosec 30˚ + 16 sec 60˚ - 8 cot 45˚ = 48

Example 5 for trig ratio table

Evaluate 7 cosec 30˚ + 16 sec 60˚ - 8 cot 45˚

Solution

7 cosec 30˚ + 16 sec 60˚ - 8 cot 45˚

7 x 2 + 16 x 2- 8

14+ 32 -8

46 - 8 = 38

7 cosec 30˚ + 16 sec 60˚ - 8 cot 45˚ = 38

Thursday, January 17, 2013

Use of Calculus in Real Life

In this article we are going to discuss about the use of calculus in real life problems step by step concept. The process of contraction the calculus comparable to over screening the real life problems concept and problem in calculus are referred as review use of calculus in real life. This article helps to improve the knowledge the real life for calculus and below the problems are using toll for the help with test calculus. Put down behind the useful get review real life use of calculus to this article. Use of calculus in real life solutions also show below.

Level One Example Real Life Calculus Use Problems:-

Real life calculus use problem 1:-

Integrate the following equation

`int 5x^4+4x^5+5x^3 dx`

Solution:-

Step 1:-

`int 5x^4+4x^5+3x^3 dx`

Step 2:-

`= int5x^4dx + int4x^5dx + int 3 x^3 dx`

Integrating the above equation

We get

Step 3:-

`= (5x^5)/(5) + (4x^6)/ (6) + (3x^4)/ (4)`

Step 4:-

`= (5x^5)/(5) + (4x^6)/(6) +(3x^4)/ (4)`

Step 5:-

`= x^5+ (2x^6)/ (3) + (3x^4)/ (4)`


Real life calculus use problem 2:-

Solve by differentiating the following equation and get the first derivative second derivative and third derivative

`y = 12x^3 + 8x^2+ 4x^5 + x`

Solution:-

Given equation is `y = 12x^3+ 8x^2 + 4x^5 + x`

Step 1:-

To get the 1st derivative differentiate the above given equation

`(dy)/(dx) = 36x^2+16x+20x^4`

Step 2:-

To get the 2nd derivative differentiate the ist derivative

`(d^2y)/ (dx^2) = 72x+16+80x^3.`


Level Two Example Real Life Calculus Use Problems:-

Real life calculus use problem 1:-

Integrate the following equation

`int 8x^4+4x^5+2x^3 dx`

Solution:-

`int 8x^4+4x^5+2x^3 dx`

Step 1:-

`= int 8x^4 dx + int 4x^5 dx + int 2 x^3 dx`

Integrating the above equation

We get

Step 2:-

`= (8x^5)/ (5) + (4x^6)/ (6) + (2x^4)/ (4)`

Step 3:-

`= (8x^5)/(5) + (2x^6)/(3)+ (x^4)/(2)`

Real life calculus use problem 2:-

Differentiate the following equation and get the 1st, 2nd and 3rd derived

`y = 8x^3+4x^2+2x^1+ 12`

Solution:-

Step 1:-

By differentiate the given equation with respect to x to get the 1st derivative

`y= dy/dx =24x^2+8x^1+2.`

Step 2:-

To get the 2nd derived differentiate the 1st derivative of the given equation.

`y = (d^2x)/ (dy^2) = 48x + 8`

Step 3:-

To get the 3rd derived differentiate the 2nd derivative of the given equation.

`y= (d^3x)/ (dy^3) = 48.`

Friday, January 11, 2013

Function of X Graph

In the mathematics concept, a polynomial function is an expression of finite measurement, lengthwise it is constructed from variables and constant values, using only the operations of addition, subtraction, multiplication, and non-negative, whole-number exponent values. For example, 3x2 − 5x + 9 is a polynomial, but 3/x2 − 5/x + 9x3/2 is not, because its second term involves division by the variable x and because its third term contains an exponent that is not a whole number. In this article we shall discuss about function of x graph.

Sample Problem for Function of X Graph:

Problem 1:

Solve the function f(x) = 2 x + 3.find the coordinate of vertex of the triangle formed by lines and the x axis in graphing polynomial.

Solution:

Given:

        F(x) = 2x + 3

We take f(x) = -y

We get,   2 x – y + 3 = 0

        Thus, 2x – y = -3

                        2x = y – 3

Equation (1) is divided by -1. We get

                              -2x = -y + 3

                              y = 2x + 3 -------------- (1)


From equation (1) we get the following values

            X         -2         -1           0        1
            Y          -1          1          3         5

Graph

Problem 2:

Solve the function f(x) = -x + 4.find the coordinate of vertex of the triangle formed by lines and the x axis in graphing polynomial.

Solution:

Given:

        F(x) = -x + 4

We take f(x) = y

We get,   -x + y + 4 = 0

        Thus, -x + y = -4

                        -x = -y – 4

                         y = x - 4

From equation (1) we get the following values

            X      0            1           2           3
            Y     -4           -3          -2           -1

Graph

Function of X Graph:problem 3:

Solve the function f(x) = -2 x - 3.find the coordinate of vertex of the triangle formed by lines and the x axis in graphing polynomial.

Solution:

Given:

        F(x) = -2x - 3

We take f(x) = -y

We get,   -2 x - y - 3 = 0

        Thus, -2x - y = 3

                          -2x = y + 3

Equation (1) is divided by -1. We get

                              2x = -y -3

                              y = -2x - 3 -------------- (1)

From equation (1) we get the following values


            X         -2         -1       0            1          
            Y          1         -1      -3           -5

Graph     

Wednesday, January 9, 2013

Quadrilateral Right Angle

Quadrilateral is a two dimensional figure which has four sides and four inside angles. In Quadrilateral, All sides are equal lengths and equal angles. It is also called as four sided polygon with four angles. Quadrilateral is a type of plane geometry. There are many quadrilaterals in geometry. They are square, rectangle, rhombus, and parallelogram, and kite, trapezoid. Let us learn about the right angle quadrilaterals.



Quadrilateral Right Angle:

rt1

An internal angle that is equal to 90° is said to be right angle.

rt2

These are various kinds of right angles.

Various properties of Quadrilateral:

  • All quadrilaterals contain 4 surfaces.
  • A square has acquired 4 surfaces of identical length and 4 right angles.
  • A Rhombus has acquired 4 sides of identical length and opposite sides is parallel and angles are identical.
  • The rectangle holds 4 right angles. It has acquired 2 pairs of identical length surface.
  • A trapezium has acquired one pair of parallel sides.
  • A kite has acquired two pairs of surface next to each other that have identical length.

Types of Quadrilaterals with Right Angles:

Square:

rt3

A square has equivalent surface and each angle is a right angle. As well as, opposite surface are parallel.

Rectangle:

rt5

Rectangle is a shape with 4 sides where each angle is a right angle. As well as, differing surface are parallel and of identical length.

Rhombus:

rt6

A rhombus is figure with 4 sides where all surface have identical length. Rhombus does not have the right angle. But the diagonal got right angles. Also differing surface are parallel and opposite angles are identical.

Kite:

rt9

It has two pair of surface. Every pair is prepared up of adjacent sides that are identical in length. The angles are equivalent where the pairs get together. Diagonals get together at a right angle, and one of the diagonal bisects another.

Trapezium:

im8

Both of these are kind of trapezium. Every of them have dissimilar properties in the quantity of right angles. But each have 4 surface and one pair of parallel surface.