Monday, August 5, 2013

Variable in Math and its Examples

We can define variable as a symbol or letter represents an unknown value. This means, the change of variable. Generally, we can show variable in math by the letters like x or y. For example, 5 x – 4 = 7, in this x is one of such. The number 5 is the coefficient of ‘x’ and 4, whereas 7 are the constant terms. We can define the kinds of these. They are Quantitative, qualitative, independent and dependant variables.
The Quantitative one means, always there is a change in numeric value. For example, Height, weight and age. Qualitative means, not having the particular order. For example, Names of persons may be peter or may be john or may be Emily or……..

Examples for solving for variables:
1. Find the value of the value of ‘ x, from the equation 3x – 7 = 9 – x.
Solution: The given equation is, 3x – 7 = 9 – x. To solve this, we can add + 7 on both sides. 3x – 7 + 7 = 9 + 7 – x which makes it 3x = 16 – x. Now, we can add the + x on both sides of the equation.      3x + x = 16 – x + x which makes it 4x = 16. Now, we can divide the equation by 4. Then the value of the given ‘ x is 4.

2. Find the value of the variable rates x, from the equation 14 – 3y = 4y.
Solution: The given equation is, 14 – 3y = 4y. To find the value of the unknown, we can Add + 3y on both sides of the equation. 14 – 3y + 3y = 4y + 3y or 14 = 7y. We can write this equation as 7y = 14. Then we can divide this equation by 7, we get y = 2. The value of the given ‘ y is 2.
3. Find the value of the value of ‘ r in the equation r – 4 + 6r = 3 + 8r.
Solution: r – 4 + 6r = 3 + 8r. First, we can add the - 8r on both sides of the equation. r – 4 + 6r – 8r = 3 + 8r – 8r. We get - r – 4 = 3. Then, we can add 4 on both sides, which makes it -r – 4 + 4 = 3 + 4 and - r = 7 and r = - 7.

Monday, July 22, 2013

Help Your Child To Do Homework

Homework means a home task in which the teacher gives the responsibility to the student to work himself or she on the topics covered in the class.

1.    First of all the parents should check the topic which is taught in the class
2.    It should be asked by the student, whether he or she has comprehended the topic or not.
3.    It is not essential that parents who have done Maths specialization can only teach maths, but it’s essential to comprehend the concept and help the child in either ways.
4.    Always try to give positive comments. It’s wrong to say in front of the child that “Don’t worry; I was bad in Maths too.” This makes a child loose interest in understanding the mathematical concepts and the child loses interest in Maths.
5.    Even if you are not aware about the topic, then just take a look at the examples which the teacher has given to the student in the classwork.
6.    Motivate the student to try again and again because to be perfect in Maths one has to practice the questions like math fractions.

1.    Activity oriented – If a child is unable to solve the problem then make the child sit and do the activity for simple learning. The same technique can be applied in completion of homework also. First let the child practice the problems in rough and then he or she can do the homework. For example if a child is not able to comprehend the concept of plus and minus in Maths. What you can do is that take one pencil in your hand and after sometime takes one more pencil. Explain the child that one and one is two, this is the concept of addition and if you will take out one pencil, then only one will remain, this is the concept of subtraction. After explaining the concept tells the child to comprehend and solve the homework in your presence and with your support.
2.    Audio visual aids – if a child is unable to solve the geometric problem, use the technology of internet. Show the child colourful pictures of square, rectangle, circle etc. After showing the picture help the child for constructing the figure using draw and paint with different colours. This will make more interesting for the child and he or she will remember for a long time. Then tell the child to attempt the questions given in the homework.
3.    Daily life experiences – If a child wants to multiply, divide or wants to calculate the amount of any object, give the child the duty to take care of the money spend on his stationary in a month. On the first day of the month tell him or her to present the bill. By this experience he will be able to calculate maximum calculations and he can use the calculations in the homework.
This is how we can talk about math homework help and the things we should take in mind when we are doing it.

Wednesday, July 10, 2013

Formula for Surface Area of Cylinder

A cylinder is one of the common shapes we see in everyday life. It is three dimensional. It has two parallel faces which are called as bases or ends of the cylinder. A curved surface joins these two bases. Therefore, a cylinder area, more predominantly referred as surface area of cylinder, includes two types of areas. One is the area of the bases, called as base areas, and the other is the area of the joining curved surface. The latter is better known as the lateral surface of a cylinder. In general surface area of cylinder refers to the sum of lateral area and the areas of the two bases.

Let us illustrate how the formula for surface area of a cylinder can be derived in a simple way.

In the above diagram, a cylinder of radius r and height h is shown on the left. Suppose the cylinder (considering it to be hollow) is ripped along the dotted line shown and opened on both the directions, we can see a rectangle formed as shown on the right. The length of the rectangle is equal to the circumference of the end circles which is 2πr, because the entire circumference is opened out The height of the rectangle remains same as the height of the cylinder as h. It can be seen as obvious that the area of the curved surface of the cylinder is same as the area of the rectangle, which is 2πr* h = 2πrh. Therefore, the lateral surface area of a cylinder is given by L = 2πrh.

The total surface area is the sum of the lateral surface area and the areas of the two bases which are circular. The areas of the two bases are the areas two circles which is πr^2 + πr^2 = 2πr^2. Therefore, the total surface area of the cylinder is, (2πr^2 + 2πrh). The same can be simplified in factored form as 2πr(r + h).

Now let us look into the significance of surface area and lateral area of a cylinder. The surface area tells us the amount of surface, in turn, the area of the sheet metal to that is needed to fabricate a cylinder. In many cases, cylindrical shapes in large sizes without ends are used as shells (think of water or any fluid pipe lines). Therefore, in such cases, only the lateral surface areas of cylinders play the role.

There are some more interesting facts about the cylinder areas and cylinder volumes. Suppose you are asked to design a hollow cylinder for a given volume that involves minimum cost to fabricate. That is, to find the dimensions with least surface area. Mathematically we can prove that for a fixed volume of a cylinder, the least surface area can be achieved when h = 2r, that is when the height of the cylinder equals the diameter of the cylinder. Same way, for a given surface area of a cylinder, maximum volume can be obtained by making the height and diameter of the cylinder congruent.

Tuesday, July 2, 2013

What is division property of equality

Equality is a property of an equation. An equations says that two expressions are equal and the symbol ‘=’ appears between those two expressions. In such cases the equality is not affected by adding or subtracting the same number on both sides. The number can also be 0 and still the equality is not disturbed.
Similarly the equality of an equation is not affected by multiplying or dividing both sides, again, by the same number. But an important point in the cases of multiplication or division the number cannot be 0. So with the restriction of 0, the equality of an equation is maintained when divided by the same number on both sides. This is what is known as division property of equality or simply as division property.

So, the division property of equality definition is, any equation maintains its equality status when divided on both sides by the same number except 0.

Let us illustrate some cases as division property of equality examples.
1) 4 = 4. When you divide by 2 on both sides you get 2 = 2 and when divided by 4 on both sides the result is 1 = 1. The answers are true in both cases. Also, the equality is not affected even when you divide by same negative numbers. For example, 4/(-2) =  4/(-2)? -2 = -2.

2) Let us try an equation involving simple variables. Say 3x = 6. When divided by 3 on both sides you get the solution as (3x)/(3) = (6)/(3) or x = 2. Same way, (3x)/(-3) = (6)/(-3) or -x = -2, which is also true.

3) Now let us discuss an example where many students tend to commit a mistake by not properly applying the division property.

Let, x2 = 4x be an equation. Now I divide both sides by x and solve x = 4. But x2 = 4x can be rewritten as a quadratic equation form as, x2- 4x = 0. A quadratic equation has always two solutions whereas we found only one solution. How and why the other solution is missing? What is the fallacy here?

The myth is, the step of dividing both sides of the equation x2 = 4xby x. Because if x = 0, the equation x2 = 4x is true and hence the variable can also take the value of 0.

When such being the case, the division by x is not valid and hence one should not have gone ahead with that step. The correct method of solving is x2 = 4x?x2 - 4x = 0?  x(x – 4) = 0. Now as per the zero product property there are two solutions as x = 0 and x = 4.

Thursday, May 16, 2013

Chain Rule

When a function is a composite of two or more functions then to find the derivative of that function, we use chain rule. Let us understand the proof of chain-rule:
Let y = f(u)
And z= g(x)
where y and z both are differentiable functions.
Then  y= f(g(x)) is a function of x
delta  y = f(u+delta  u ) – f(u)
And  delta  z = g(x+delta  x) – g(x)
Therefore delta  y/ delta  z = f(u+delta  u ) – f(u)/delta  t
And delta  z/ delta  x = g(x+ delta  x) – g(x)/delta x
We first suppose that for sufficiently small values of delta  x , delta  z not equal to 0. This will be the case if dz/dx not equal to 0. Then, g is differentiable, it is continuous and hence when delta  x ->0, x+delta x->x and g(x+delta x)->g(x).
Therefore z+delta z->z and delta z->0. Since delta z not equal to 0 as delta x->0, we may write
delta y/delta x= delta y/delta u . delta z/delta x
We can see that f and g, both are continuous functions as they are differentiable. Thus, delta u->0 when delta x->0 and delta y->0 when delta u->0.
Therefore, limit delta x->0 delta y/delta x = limit delta z->0 delta y/delta z multiply limit delta x->0 delta t/delta x
Hence dy/dx = dy/dz. dz/dx

For example, find derivative of sin(x^2)
Solution: Here we can clearly see that the function is function of a function.
Suppose y = sin(x^2) = sin z
Where z = x^2
Therefore dy/dz = cos z and dz/dx = 2x
Hence by this rule
Dy/dx = dy/dz. dz/dx
          = cosz . 2x            [ substitute the values of dy/dz and dz/dx]
         = 2xcos(x^2)         [ substitute the value of z]

Derivatives chain rule: Sometimes it is also known as chain-rule derivatives.  It means to find derivative of a function using this rule.

Reverse Chain Rule:
It is used for integration or say anti derivative of functions. It is equivalent to integration by parts.

The Chain Rule Multivariable:
It means many variables are included in this form. Suppose that y=f (a, b), where a’ and b themselves depend on one or more variables. This Rule gives us the liberty to differentiate y with respect to any of the variables (a, b) involved:
Let a=a (t) and b=b (t) be differentiable at t and let y=f (a, b) is differentiable at the point (a (t), b (t)). Then y=f (a (t), b (t)) is differentiable at t and
  Dy/dt = dy/ da . da/dt + dy/ db . db/dt
Let us understand an example of the multivariate chain-rule:
For example: Solve it using the multivariate chain-rule:  Let y = a^2b-b^2 where a and b are parameterized as a=t^2 and b=2t
Then Dy/dt  = dy/ da . da/dt + dy/ db . db/dt
                     =2ab.2t +(a^2-2b)(2)
                     = (2t^2. 2t).2t+((t^2)^2 – 2(2t))(2)
                     = 8t^4 + 2t^4 – 8t
                     = 10t^4 -8t