Radicals and Exponents

The opposite operation to the exponent is known as radical. A radical is an expression which contains the square roots, cube roots etc. For example the expression √49 can also be called as square root of 49 or root of 49. Radicals have the same property of the numbers. The simplification is to reduce the numbers and reduce the power of variables inside the roots.

Exponent of a number shows you how many times the number is to be used in a multiplication.  It is shown as a small number to the right and above the base number.  In this example: 22 = 2 × 2 = 4 ( Another name for exponent is index or power ).

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Examples for Radicals:

Example 1 for radical problem:

Simplify the radical:    √ (144) / (√36)

Step 1: Factors of 144 = 12 × 12

√ (144) = √ (12 × 12)

= √122

Step 2: Square root of 122 = 12

Step 3: Factors of 36 = 6 × 6

√ (36) = √ (6 × 6) = √62

Step 4:  √62 = 6

Step 5: so, √ (144) / √36) = 12 / 6

The answer of the given radical is 12 / 6
Example 2 for radical problem:

Simplify the radical:    3√8 × √12

Step 1: Find the factor of 8 = 2 × 2 × 2

Step 2: Take cubic root of 8 = 2

Step 3: Find the factor for 12 = 2 × 2 × 3

Step 4: Take square root of 12 = 2 √3

Step 5: so, 3√8 × √12 = 2 × 2 √3

The answer for the given radical is = 4√3

Examples for exponents:

EXAMPLE-1:

453 . 454 = 453+4 = 457

EXAMPLE-2:

148 / 145  = 14 8-5 = 143

EXAMPLE-3:

( 92 )3  = 9 2x3  = 96

EXAMPLE-4:

( 11 / 13 )2  = 112 / 122

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EXAMPLE-5:

70 = 1

EXAMPLE-6:

5-3 = 1/ 53

Inverse Variation

To understand the inverse variation,let is correlate it with our real life of inverse variation examples such as the total time taken with the speed traveled in a trip. Assuming distance of a trip would be fixed, let us say the distance of a trip would be 250 mile.

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Here  inverse variation equation will be used as T=250/r , according to the time will be  T be time of the trip will be equal to 250  divided by the Rate in miles per hour which will called r .so if rate is 50 miles per hour it would take,  5 hours  .

The next of the inverse example is total time taken to spread the soil and the number of people working on a part of land. so let us say if the total time of the job will take 12 hours the total time T to complete the job would be  12 hours divided by n, ( n is the number of workers).

Here the equation will be T= 12/n.  so if have 1 worker , we take 12 hours, 3 workers we  take 4 hours and so on.

Total amount of cost required by per person for petrol and the number of people in a cab. So if we know that it will cost 45 dollars’ worth of petrol to go for a picnic trip. The cost represents C= 45 dollars divided by number of people in the cab.

That is C= 45/no of people. So if n will increases the total cost decreases.
Here in inverse variation definition is as the ones quantityvalue gets bigger and the others value gets smaller in such that there product remains the same or proportional. Let us understand with one more example, Y varies inversely as X. where Y= 4 and when X=2.

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Find the value of the inverse-variation equation. Then determine Y when X= 16.  By performing substitution we can solve this question.  Y= k/x.  by doing cross multiplication , we get 4= k/2 which is k=8.  So our inverse-variation will be  Y=K/x that is 8/x. Now we can answer the second query on determining the y when x is 16. So, y= 8/x , substituting the value of x as 16. So y= 8/16. And we get  1/2 . y= ½ and x= 16.

Solve Laplace Equation

In mathematics, Laplace equation is one of the most interesting topics in second order partial differential equations. The two dimensional heat equations in Cartesian form in unsteady state is

`(delu)/(delt)` = `alpha^2 [(del^2u)/(delx^2) + (del^2u)/(dely^2)]`

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For the steady state `(delu)/(delt)` = 0, then the above equation becomes

`[(del^2u)/(delx^2) + (del^2u)/(dely^2)]` = 0

(or)

`grad^2 u` = 0

This is the two dimensional heat equation in unsteady state (or) Laplace equation. In this article, we shall discuss about how to solve the Laplace equation. The following are the example problem in solve Laplace equation.

Solve Laplace equation - Laplace equation:

Finite difference solution for the two dimensional heat equation in steady state (or) Laplace equation is

`grad^2 u` = 0

(or)

`(del^2u)/(delx^2)` + `(del^2u)/(dely^2)` = 0

Here we know the values of `(del^2u)/(delx^2)` , `(del^2u/dely^2)`

`(del^2u)/(delx^2)` = `(u_(i+1, j) + u_(i-1, j)-2u_(ij))/(h^2)`

`(del^2u)/(dely^2)` = `(u_(i, j+1) + u_(i, j-1)-2u_(ij))/(h^2)`

substitute the values of `(del^2u)/(delx^2)` , `(del^2u)/(dely^2)` in the Laplace equation, then we get

`4u_(ij)` = `(u_(i+1, j) + u_(i-1, j) + u_(i, j+1) + u_(i, j-1))`

`u_(ij)` = (`1/4` ) `(u_(i+1, j) + u_(i-1, j) + u_(i, j+1) + u_(i, j-1))`

This is the finite difference solution for the Laplace equation, which is also called as the standard five-point formula.

Solve Laplace equation

Solve Laplace equation - Example problem:

Example 1:

Solve the laplace equation for the given region

Solve Laplace equation - Example

Solution:

The finite difference sheme for the Laplace equation is

Solve Laplace equation - Example

u(ij) = (`1/4` ) [`(u_(i+1, j) + u_(i-1, j) + u_(i, j+1) + u_(i, j-1))` ]

Using this we can solve this,

u1 = `1/4` [200 + 50 + u2 + u4]

= `1/4` [250 + u2 + u4]                                → (1)

u2 = `1/4` [u1 + 0 + 0 + u3]

= `1/4` [u1 + u3]                                         → (2)

u3 = `1/4` [u4 + u2 + 100 + 50]

= `1/4` [150 + u2 + u4]                                → (3)

u4 = `1/4` [100 + 200 + u1 +  u3]

= `1/4` [300 + u1 + u3]                                → (4)

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Solve the equation  by using the gauss seidal method

u1 = `1/4` [250 + u2 + u4]

u2 = `1/4` [u1 + u3]

u3 = `1/4` [150 + u2 + u4]

u4 = `1/4` [300 + u1 + u3]

Let the intial approximation is

Learn Slope Form

The ratio of rise value of the line equation to the run value of the line equation is called as slope. Slope form learning is one of the important part of algebra. In other words, it can be defined as the ratio of the change of  Y values to the change of the X value. Generally, the slope of the equation is denoted as "m". It is necessary to learn the slope form. learn slope form is mainly used in linear equation.

learn slope form of the linear equation

The slope form of the linear equation can be written as,

y = mx + b

Where, y is the line equation,

m is the slope of the equation and

b is the Y intercept value.

In point form, the line equation can be written as,

Y - Y1 = m (X - X1)

Where, X1 and y1 are the points of the line

m is the slope of the equation.

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Given two points, we could find the slope using the formula

m = (Y2 - Y1) / (X2 - X1)

Where, (X1, Y1 ) and  (X2, Y2 ) are any two points in the line.

Example problems for learn slope form

Ex:1 Find the slope of the given equation Y = 7X + 6.

Sol: In general form, the line equation is given as,

y = mx + b

where, m is the slope of the equation.

In the above equation, the slope is coefficient of x,

slope = 7

Ex:2 Find the slope of the given points (1, 3) and (3, 6).

Sol:

The given points are (1, 3) and (3, 6).

Here, X1 = 1, Y1 =3

X2 = 3, Y2 = 6

The formula for finding the slope is given as,

m = (Y2 - Y1) / (X2 - X1)

Substitute all  values in the above equation,

m = (6 – 3) / (3 -1)

= 3 / 2

Ans:

The slope of the given equation is m= 3/2

Ex:3 Find the slope of the given points (2, 6) and (6, 14).

Sol: The given points are (2, 6) and (6, 16)

Here, X1 = 2, Y1 = 6

X2 = 6, Y2 = 16

The formula for finding the slope is given as,

m = (Y2 - Y1) / (X2 - X1)

Substitute all  values in the above equation,

m = (16 – 6) / (6 -2)

= 5 / 2

Ans:

The slope of the given equation is m = 5 / 2

Expressions With Rational Exponents

A rational number is a number represented as p/q where p and q are integers and q != 0.Exponent means  power.

The large numbers are very difficult to read and write and understand. To make them simpler we can use exponents, with the help of this many of the large numbers are converted to simpler forms. The short notation 74 stands for the product 7 x 7 x 7x 7. Here ‘7’ is called the base and ‘4’ the exponents. The number 74 is read as 7 raised to the power of 4 or simply as fourth power of 4. 74 are called the exponential form of 2401.

The above example is of a natural number as exponents.

Ex 5 ^(3/2) here 3/2 is a rational number so this is an example of rational exponents.

Rules invovled for solving rational exponents:

The Rational exponents rules are:

p,q are any real numbers except zero and m,n are positive integers.

Rule I:The base value is same under multiplication so we can directly add the power values. Then the exponent take the following form

pm  × pn = pm+n

Solving the expression 73  × 75 = 73+5 =78

Rule II:The base value is same under division so we can directly subtract the power values. Then the exponent take the following form such that m > n then

(p^m)/(p^n) = pm-n

Solving the expression  (4^4)/(4^2)= 44-2=42

Rule III:If m < n, (p^m)/(p^n) =(1)/(p^(n-m))

Solving the expression 8^3/8^2 =1/8

Rule IV:A real number to the power of power

(pm )n  = pm*n

Solving (82 )3  = 82*3=86

Rule V:The power of 0 is

p0 = 1. Anything power zero is equals to 1.

p^m/p^m   = pm-m =p0 = 1

Rule VI:p to the power of negative number is

p-m =1/p^m
Solving the expression 2-5 =  1/2^5

Rule VII:Two numbers to the same power can be written as

pm  × qm = (p*q)m

Solving the expression 32 × 82 = (3*8)2=242

Rule VIII A number to rational power is written as

p^(m/n) = root(n)(p^m)

Examples of expressions having rational exponents:

Ex : 1Solve the exponent    2x^(1/3)xx8^(5/3)xxx^(2/3)

Sol:Step 1:Given

2x^(1/3)xx8^(5/3) xx x^(2/3)

Step 2:Exponents are added

=2x^(1/3+2/3) (2^3)^(5/3)

Step 3: 8 is written as 2 power 3

Step 4 : Simplify=2x2^5

=2xx32 x

=64 x

Ex 2: Solve x^(-1/3)8^(-2/3)

x^(-1/3)8^(-2/3)

Solu: Step 1  x^(-1/3) 8^(-2/3)

Step2: 8^(-2/3) = (2^3)^(-2/3) = 2^(3 xx-2/3) = 2^-2 = 1/2^2 = 1/4

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Step3 Simplify the variable with negative exponent

=1/(x^(1/3)) 1/4

Step 4 Simplify:=1/(4x^(1/3))

=1/(4root(3)(x))

Power Coefficient Equation

A coefficient is a multiplicative factor in a few phrase of an expression. It is typically a number, but in any case does not occupy some variables of the equation. For example in equation 7x2 − 3xy + 1.5 + y the first three stipulations correspondingly have coefficients 7, −3, and 1.5. In the third expression there are no variables, so the coefficient is the idiom itself called the power constant term or constant power coefficient of this expression

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Example Problem for the Coefficient:

Find the coefficient of the given equation:

X2 + 5y3 +2xy2 + 5yx2

Solution:

Given that X2 + 5Y3+ 2XY2 + 5YX2

Here X2+ 5Y3+ 2XY2+ 5YX2 the expression with the coefficient

The coefficient of x2 is 1

The coefficient of  5Y3 is 5

The coefficient of 2xy2 is 2

The coefficient of 5yx2 is 5.

Coefficient Of Varience

Coefficient of variation is a relative determine for standard deviation.

It is defined as 100 times the coefficient of spreading base in the lead standard difference is called coefficient power of variation.

Less C.V. designate the fewer variability or additional power.

More C.V. indicate the more variability or less power.

According to Karl Pearson who suggested this compute the equation, C.V. is the percentage dissimilarity in the mean, standard deviation being considered as the total variation in the mean.

With the help of C.V. we can find which salesman is more consistent in making sales, which batsman is more consistent in scoring runs, which student is more consistent in scoring marks, which worker is more consistent in production.

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Coefficients Of Dispersion

Whenever we want to compare the variability of two equations which differ widely in their averages or which are measured in different units.

We do not merely compute the measures of dispersions in the equations but we calculate the coefficients of dispersion which are pure numbers autonomous of the units of measurement.

Quotient Rule for Integration

The quotient rule is a method of finding the integration of a function that is the quotient of two other functions for which derivatives exist.

By the Product Rule,

if f (x) and g(x) are differentiable functions, then
d/dx[f (x)g(x)]= f (x)g'(x) + g(x) f' (x).
Integrating on both sides of this equation,

∫[f (x)g'(x) + g(x) f '(x)]dx = f (x)g(x),
which may be rearranged to obtain

∫f (x)g'(x) dx = f (x)g(x) −∫g(x) f' (x) dx.     (A)
Letting U = f (x) and V = g(x)

then differentiating it  we get

dU = f '(x) dx and dV =g'(x) dx,

pluging these values in (A), we get

∫U dV = UV −∫V dU.                  (1).


By the Quotient Rule,

if f (x) and g(x) are differentiable functions, then

d/dx[f (x)/g(x)]= g(x) f '(x) − f (x)g'(x)/[g(x)]2 .
Integrating both sides of this equation, we get
[f (x)/g(x)]=∫g(x) f '(x) − f (x)g'(x)/[g(x)]2 dx.
That is,

f (x)/g(x)=∫f '(x)/g(x)dx -∫f (x)g'(x)/[g(x)]2 dx,
which may be rearranged to obtain

∫f '(x)g(x)dx = f (x)g(x)+∫f (x)g'(x)/[g(x)]2 dx.      (B)

Letting u = g(x) and v = f (x) and then differentiating it , we get

du = g'(x) dx and

dv = f '(x) dx,
we obtain a Quotient Rule Integration by Parts formula:

plugging values of u , v , du and dv in B we get
∫dv/u= v/u+∫(v/u²) du.                                  (2)

quotient rule for integration-Application

∫[sin(x−½)/x²] dx.
Let
u = x½, du=1/2(x-½)

v=2cos(x-½),dv = sin(x−1/2)/x³/²

Then

∫sin(x−½)/x² dx = 2 cos(x-½)/x½+∫2 cos(x-½)/x• 1/2(x-½) dx

= 2 cos(x-½)/x½− ∫2cos(x-½) ·•[−(x-³/²)/2]dx

= 2 cos(x-½)/x½− 2 sin(x-½) + C,
which may be easily verified as correct.

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quotient rule for integration -Illustration


The Quotient Rule Integration by Parts formula (2) results from applying the
standard Integration by Parts formula (1) to the integral

Let ∫dv/u

with

U = 1/u,V = v,then differentiating ,we get

dU = − 1/u² du,dV=dv
plugging these values

∫dv/u=∫U dV

= UV −∫V dU

= 1/u·( v) −∫v (− 1/u²)du
= v/u+∫v/u² du