Thursday, August 20, 2009

How to find critical points for an equation

In calculus, a critical point of a function of a real variable is any value in the domain where either the function is not differentiable or its derivative is 0. The value of the function at a critical is a critical value of the function. These definitions admit generalizations to functions of several variables, differentiable maps between Rm and Rn, and differentiable maps between differentiable manifolds.

Lets see a problem from calculus help with probability problems,which explains us
more about this .

Question:-

f(x)=3x4-4x3-12x2+6

Differentiating with respect to x

Solution:-

f'(x)=12x3-12x2-24x

we know f'(x)=0 ,so

12x3-12x2-24x=0

12x(x2-x-2)=0

12x(x2-2x+x-2)=0

12x(x-2)(x+1) = 0

x(x-2)(x+1) = 0

Now we use the zero product property ,which states that
if ab=0 ,then a=0 and b=0,this property is from geometry terms and definitions .

so x=0 (x-2)=0 (x+1)=0

x=0 x=2 x=-1

Therefore,x=0,-1,2 are the critical points for y=f(x)

Wednesday, August 19, 2009

Simple Algebraic Equation Problem

Topic: Algebraic Equation

An equation is a mathematical statement, in symbols, that two things are exactly the same . Equations are written with an equal sign, as in
2 + 3 = 5
9 - 2 = 7
The equations above are examples of an equality: a proposition which states that two constants are equal. Equalities may be true or false
How to solve Equations ?
Its simple.Here is a example of 7th grade math equations :


Thursday, August 13, 2009

Solving a integral problem by using substitution method

Topic:- Integration

Integration is an important concept together with differentiation, forms one of the main operations in calculus help
Given a function ƒ of a real variable x and an interval [a, b] of the real line, the definite integral
\int_a^b f(x)\,dx \, ,
is defined informally to be the net signed area of the region in the xy-plane bounded by the graph of ƒ, the x-axis, and the vertical lines x = a and x = b.

Let's work out a simple example on this.
Question:-

solve ∫ ( x / √x2-9 )

Answer:-

We do it by substitution method.
 Let u = x2-9
    
    du
    ---- = 2x
    dx
 
   du
or ---- = xdx
   2
substituting these values ,the integral becomes

∫ (du/2) / √u

= 1/2 ∫ (u)-1/2 du

        (u)-1/2 + 1
= 1/2 ---------------
         -1/2 + 1
by equalizing the denominators
   
    -1+2
   ------- = 1/2
      2
So the integral becomes
    
        (u)1/2
= 1/2 ---------------   + c
         1/2

= (x2-9)1/2+C (as u =x2-9)

= √( x2-9 ) +c  is the Answer

For more help on this ,you can reply me .

Wednesday, August 12, 2009

A problem on Integrating Factor

Topic:Integrating Factor

In simple term integrating factor in mathematics, is an function that is chosen to facilitate
the solving of a given equation involving differentials. It is generally used to solve ordinary differential equations, but is also used within multivitamin calculus,in this case often multiplying through by an integrating factor allows an inexact differential to be made into an exact differential (which can then be integrated to give a scalar field).
In simple term integrating factor is denoted as I.F.
Question:
   
           dy
solve     ____ + = ex
           dx

Answer:

   Standard form: dy
                 ____ + p(x) * y = q(x)    Here p(x) = 1
                  dx                            q(x) = ex


 Now Inergrating factor = e√pdx = e√idx = ex

 Now the equation is:

                     y = i
                        ___
                        I.F  √I.F * Q(x) dx + c

Plugging in all the values:

                     y = 1
                        ___
                         ex √(ex * ex) dx + c

                     y = 1
                        ___
                         ex √(ex)2 dx + c

                     y = 1 * e2x
                        ___ ___ + c
                        (ex) 2

                     y = ex
                        ___ + c
                         2        


For help you can contact us

Friday, July 17, 2009

a problem on quadrilaterals

Topic:-quadrilateral

In geometry, a quadrilateral is a polygon with four 'sides' or edges and four vertices or corners. Sometimes, the term quadrangle is used, for analogy with triangle, and sometimes tetragon for consistency with pentagon (5-sided), hexagon (6-sided) and so on. The word quadrilateral is made of the words quad and lateral . Quad means four and lateral means sides. The interior angles of a quadrilateral add up to 360 degrees of arc.

Question:-

In a quadrilateral, the third angle is 2 times the fourth and the second angle is 3 times the fourth. The first angle is given as 120. Find the fourth angle.

Solution:-

Let the fourth angle be 'x'.

Here, the third angle = 2 times the fourth angle = 2x.

The second angle = 3 times the fourth angle = 3x.

We know the sum of the angles of a quadrilateral = 360°.

Therefore x + 2x + 3x + 120 = 360.

ie 6x + 120 = 360.

ie 6x = 360 - 120 = 240.

ie x = 240 / 6 = 40.

Hence the fourth angle = 40°.

For more help on this ,you can reply me

Monday, July 6, 2009

Solving Linear Equations

Topic:- Linear Equation


A mathematical expression that has an equal sign and linear expressions is called a Linear equation.

A number that you don't know,often represented by "x" or "y" is called a variable.

We solving linear equations by clicking and dragging a number to the "other" side of the equal sign. Remember that you are "isolating" the unknown "X" to solve the problem.
(Example is provided below.)

Question:-

Solve 2x-18=15
Answer:-

This math help is a part of linear equations.

2x-18=15

Add 18 to both sides

2x-18+18 = 15 + 18

2x = 15+18

2x = 33


divide by 2 on both sides


2x      33
----- = -------
2        2


  x = 16.5

16.5 is the Answer.

Wednesday, June 24, 2009

How to find volume of a cylinder

Topic :- Volume of cylinder

A cylinder is one of the most basic curvilinear geometric shape, the surface formed by the points at a fixed distance from a given straight line, the axis of the cylinder.

Here is one sample problem on how you can find the volume.

Question:-
In a cylinder radius is 3 inches and height is 8 inches ,find the volume of the cylinder.

Solution:-

Formula for cylinder volume= Π r2 h


So Π =3.14

r =radius=3

h =height=8

Hence, Volume of the cylinder

= 3.14 * (3)2*8

= 3.14 * 9 * 8

= 226.08


So the volume of the cylinder is
226.08 cubic inches.

For more similar help you can reply me .