Derivative of Quadratic Functions

Introduction to derivative of quadratic functions:

In calculus the derivative is a measure of how a function changes as its input changes. The derivative of a function at a chosen input value describes the best linear approximation of the function near that input value. A quadratic equation is a polynomial equation of the second degree. The general form is ax2 + bx + c . The constants a, b, and c, are called respectively, the quadratic coefficient, the linear coefficient and the constant term or free term.

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Derivative Formulas:

1.  ` d / dx` (x n ) = n xn-1

2.  ` d/dx` (ex)  = ex

3.  ` d/dx` `(uv)` = `u(dv)/(dx)` + `v(du)/(dx)`
4. .`d/dx` `(u/v)` = `(v(du)/(dx) - u(dv)/(dx))/ v^2`

5.  `d/dx` (sin x) = cos x

6.  `d/dx ` (cos x) = -sin x

7.  `d/dx` (tan x) = sec2x

Derivatives of Quadratic Function Problems:

Derivatives of quadratic function problem 1:

Find the derivative of quadratic function (2x3 - 3x4 + x2) with respect to x.

Solution:

Given Quadratic function is  (2x3 - 3x4 + x2)

` d/dx` (2x3 - 3x4 + x2)  = `d/dx` (2x3) - ` d/dx` (3x4) + ` d/dx ` (x2) .

=  `d/dx` (2x3) - `d/dx` (3x4)  + `d/dx` (x2).

= (2 × 3) x(3-1) - (3 × 4) x(4-1) - 2x(2-1)

= 6 x2 - 12 x3 + 2 x

`d/dx` (2x3 - 3x4 + x2) =  6 x2 - 12 x3 + 2 x

Answer: The derivative of given quadratic function is  `d/dx` (2x3 - 3x4 + x2) =  6x2 - 12 x3 + 2 x

Derivatives of quadratic function problem 2:

Find the derivative of quadratic function  y = 3t2 + t3 - 30t. with respect to 't'.Is this topic free algebra help hard for you? Watch out for my coming posts.

Solution:

Given Quadratic function is y = 3t2 + t3 - 30t.

dy  = (3 × 2)t(2 - 1) dt + (1 × 3)t(3 - 1) dt - 30t(1 - 1) dt.

= 6t dt + 3t2 dt - 30dt.

= (6t + 3t2 - 30) dt

`dy/(dt)` = (6t + 3t2 - 30)

Answer:  The derivative of given quadratic function is   `d/dt` (3 t2 + t3 - 30t) =  3t2 + 6t - 30.

Derivatives of quadratic function problem 3:

Determine the second derivative of given quadratic function f(x) = 15x2 + 10x + 5 with respect to x.

Solution:

Given quadratic function is, f(x) = 15x2 + 10x + 5

The First derivative of quadratic function is f'

f' =` (df )/ (dx)`  = (15 × 2)x(2 - 1) + (10 × 1)x (1-1)

= 30x + 10

The Second derivative of quadratic function is, f''

` d^2/dx` f(x) = `d^2/dx` ( 15x2 + 10x + 5)

= `d/dx` (f')

= `d/dx` ( 30x + 10)

= ` d/dx` (30x) +` d/dx` (10)

= 30 `d/dx` (x) + `d/dx` (10)

= 30 + 0

= 30

Answer: The Second derivative of given quadratic function is 30

Sides of a Polygon Formula

Introduction to sides of a polygon formula:

In geometry, polygon is two dimensional shapes. It has more than two sides. All sides are straight and connected to another side. The number of vertices is equal to number of sides. Vertices are nothing but corner points of the shape. In this article we shall see how to calculate the area of regular polygon.

Sides of a Polygon Formula - Formulas:

Triangle:

Formula to find area of triangle:

Area of triangle (A) = `1/2` (b x h) square units

b – Base

h – Height

Square:



Formula to find area of square:

Area of square (A) =a2 square units

a – side length                            

Pentagon:



Formula to find the pentagon:

Area of the pentagon (A) = t2 1.72 square units

t – Side length

Hexagon:



Formula to find area of hexagon:

Area (A) = t2 2.6 square units.

t - Side length

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Sides of a Polygon Formula – Example Problems:

1. Find the area of triangle whose base is 40 cm and height is 25 cm.

Solution:

Given:

Base (b) = 40 cm

Height (h) = 25 cm

Substitute the given value in the formula

Area of the triangle (A) = `1/2` (b x h) square units

= `1/2` (40 x 25)

= `1/2` (1000)

= 500

Area of the triangle (A) = 500 cm2

2. Find the area of the square, whose side length is 27 cm

Solution:

Given:

Side (a) = 27 cm

Substitute the given value in the formula

Area of square = a2 square units

= 27 x 27

Area of square = 729 cm

3. The side length of pentagon is 13.5 cm. Find the area and perimeter of the pentagon.

Solution:

Given:

Side length (t) = 13.5 cm

Substitute the given value in the formula

Area of the regular pentagon (A) = 1.72 t2 square units

= 1.72 x 13.52

= 1.72 x 182.25

= 313.47

Area of the regular pentagon (A) = 313.47 square units

4. The side length of hexagon is 14.7 cm. find the area of the hexagon.

Solution:

Given:

Side length (t) = 14.7 cm

Substitute the given value in the formula

Formula:

Area of the hexagon (A) = t2 2.6 square units

= 14.72 x 2.6

= 216.09 x 2.6

= 561.834

Area of the hexagon = 561.834 cm2

Perpendicular Line Segments

Introduction to perpendicular line segments:

Perpendicular line:

In geometry, two line segments or planes (or a line and a plane), are considered perpendicular (or orthogonal) to each other if they form congruent adjacent angles (a T-shape).  Thus, referring to Figure 1, the line AB is the perpendicular to CD through the point B.



Fig(i) Perpendicular line

Line segments:

The line segments are the the part of the line that has two end points.It include all the points between its end points.For example,consider the following figure,

Fig(ii) Line segment

In the above figure AB is a line and CD is a line segment.C and D are the two end points of a line segment.We are going to see about the perpendicular line and line segments.

Examples for Perpendicular Line Segments:

The following are the geometry examples for line segments,

Sides of the triangle

Sides of the Rectangle

Sides of the Square

Properties of perpendicular line:

The multiplication of slopes of the perpendicular lines is equal to -1.

That is m1=slope of perpendicular line 1

m2 = Slope of perpendicular line 2

m1 × m2 = -1

Distance formula for line segments:

The following formula used to find the lengh of the line segments ,

length = `sqrt((x2-x1)^2+(y2-y1)^2))`

Here (x1,y1) and (x2,y2) are the two end points of the line segments.

Problems on Perpendicular Line Segments:

Problem 1:

Find the equation of a line which is perpendicular to 4y - x = 20 and passes through the  point (2, -3).

Solution:

Given 4y - x = 20 and the point (2,-3)

To find the perpendicular line we need to find the slope.

To find the slope we need to change the given equation into slope intercept form.

4y - x = 20

Add x on both side,

4y - x  = 20

+ x =  +x

4y = x + 20

Divide by 4 on both side,

y = (x/4) + 5

y = mx + b form

So the slope m = 1/4

We know that multiple of slopes of the perpendicular lines equals to -1

1/4 * m = -1

Multiply by 4 on both sides,

m = -4

The line equation is,

(y-y1) = m(x-x1)

(y - (-3)) = -4(x-2)

y+4 = -4x+8

Subtract  4 on both sides,

y = -4x +4

Answer : The line perpendicular to the given line is    y = -4x +4

Problem 2:

By using the line segments distance formula find the length of the line segmen xy with x(5 ,3) y(1,2)

Solution:

length = `sqrt((x2-x1)^2+(y2-y1)^2))`

x1 = 5    y1 =3    x2 = 1    y2 = 2

=` sqrt( ( 1-5)^2 + ( 2-3)^2)`

=` sqrt( (-4)^2 + (-1)^2)`

= `sqrt (16+1)`

= `sqrt (17)`

Answer :The length of the line segment xy = `sqrt (17)`

Solving Vector Cartesian Coordinates

Introduction on solving vector Cartesian coordinates:

This article is about solving vector Cartesian coordinates. Solving vector Cartesian coordinates is very simple. The tutors of tutor vista helps the students in solving vector Cartesian coordinate any time. The vector coordinates are x,y and z. Three simple methods are there to describe a vector. Directions, specific lengths, angles and projections or components are used to describe vectors. The simplest method of these is Cartesian or rectangular co-ordinate system. Below we can see about solving vector Cartesian coordinates.

Solving Vector Cartesian Coordinates

In cartesian coordinate system three co-ordinate axes x,y and z are mutually right angles to each other. Consider a point P(x,y,z) in space at a distance r from the orgin. The vector r can be represented as

r = `xbara_x``+ y bara_y + z bara_z`

Where `bara_x, bara_y,` and `bara_z` are unit vectors

x,y and z are the components vectors. Components vectors have a magnitude and direction. Unit vectors have unit magnitude and directed along the co-ordinate axis.

A unit vector in a given direction is a vector in that direction divided by its magnitude. It is given by

`a_r = r/|r|`

`a_r = (xbara_x + ybara_y + z bara_z)/sqrt (x^2 + y^2 + z^2)`

Consider the points P(x,y,z)and Q(x + dx, y +dy, z+dz) in rectangul;ar co-ordinate system. The differential length dl from P to Q is the diagonal of the parallel piped is given by

dl = sqrt ((dx)^2 + (dy)^2 + (dz)^2)

The differential area ds = dx dy

= dy dz

= dz dx

The differential volume dv = dx dy dz

Solving Vector Cartesian Coordinates

Conversion of cylindrical to cartesian system

The cylinderical co-ordinates (?, F, z) can be converted into cartesian co-ordinates ( x, y, z)

Given                                                       Transform

?                                                             x = r cos ?

f                                                            y = r sin f

z                                                             z = z

conversion of spherical to cartesian co-ordinates system

The spherical co-ordinates (r, `theta` , f) can be transformed into cartesian co-ordinates (x, y, z)

Given                                                       Transform

r                                                             x = r sin ?. cosf

?                                                            y = r sin ? sin f

f                                                            z =r cos ?


Solving Vector Cartesian Coordinates

Example Problem: Give the Cartesian co-ordinates of a point whose cylinderical are ? = 1, f = 45°, z =2.

Given

Cylinderical co-ordinates are ? = 1, f = 45°, z =2.

Cartesian co-ordinates are x, y, z

x = ? cosf

= 1.cos 45 = 0.707

y = ? sinf

= 1. sin 45 = 0.707

z = z

= 2

Cartesian co-ordinates are (0.707, 0.707, 2)

Regular Irregular Polygons

Introduction for Regular irregular polygons:

Regular polygons:

When each side and angles of any polygon are similar, then it is said to be a regular polygon. A regular polygon is one with its entire sides similar and each of its angles similar, for instance: A regular polygon with 3 surfaces is said to be equilateral triangle.

Irregular Polygons:

When angles of the any polygon are not the same, then it is called as irregular polygon.

Regular Irregular Polygons-angles of Regular Polygons:

The regular polygons’ angle is grouped into two types. They are

Exterior angle
Interior angle
Interior angle of Regular Polygons:

Angle of regular polygons = [(n - 2) / n] × 180°

where n =number of sides.

Example 1:

Octagon is an eight sided polygon

Regular Polygons’ Interior Angle = [(n - 2) / n] × 180°

Here n = 8 for octagon

So the interior angle of an octagon = [(8 – 2)/ 8] * 180°

= (3 / 4) * 180°

Interior angle of octagon = 540° / 4 = 135°

Exterior angle of Regular Polygons:

360° is the exterior angle of polygons.  Hence the exterior angle can be estimated by the formula,

Regular Polygons’ Exterior angle = 360° / n

where n is the number of sides.

Example 2:

Pentagon is a polygon with five sides. Polygon’s Exterior angle = 360° / n

For Pentagon, n = 5

So the exterior angle of a pentagon = 360 / 5 = 72°.

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Regular Irregular Polygons-area of Irregular Polygons:

The steps for determining the area of irregular polygon are,

Convert every vertex into co-ordinate in graph.
Take any line segment and down to the x - axis.
Compute the average of two altitudes.
Compute the difference of two widths.
Multiply altitude with width.
Go over steps 2 to 5 until manipulating every line segment.
Then add all the areas of every line segment.
So, we will result in area of irregular shape.

Symmetry for Irregular Polygon:

Since the irregular polygons do not have equal angles, they do not posses line of symmetry..

Addition Property of Order

Introduction of Addition Property of Order:

Let now explain about the addition property with example.

Example: c < d, the expression shows that the number c is less than the number. When add the number ‘’a’’ on the both side of the expression, the relation does not change.

C + a < d + a; Thus by using addition property we can order the number from smallest to largest or vice versa (ascending or descending order) otherwise to find the number is equality or inequality.

Example Problem – Addition Property of Order:

Example 1:

Solve the expression by using addition property of order X – 5 < 17 and check the expression by getting the value of X.

Solution:

Given: The expression is X – 5 < 17

Step 1: By using addition property we can add the number 5 on both sides of the expression, we get

X – 5 + 5 < 17 + 5

Step 2: Opposites sign of the number 5 to be canceling each other.

Step 3: So we get X is less than 22

X < 22

Check the expression X – 5 < 17 by putting the value of X = 22

Step 1: The given expression X – 5 < 17

Case 1:

If the value of X = 22, then the expression get equality form such as 22 – 15 = 17

Case 2: We get the X value as less than 22, so the value of X is 21 or less than 21 (may be up to negative infinity)

Let we put the X value is 21, we get

21 – 5 < 17

16 < 17, thus we proved that the number 16 is less than 17.

Example 2:

Prove by using then addition property of the number 7 is greater than 5.

Solution:

Given: 7 > 5

Step 1: First we add any number; let take the number 2 is added to the left hand side of the expression.

7 +2 = 9

Step 2: Then we add the number to the right hand side of the expression, we get

5 + 2 = 7

Step 3: Compare the result of the above two steps.

Thus the number 9 is greater than 7; 9 > 7

Therefore we proved that the number 7 is greater than 5

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Practice Problems – Addition Property of Order:

1. Solve the expression by using addition property of order X – 5 < 15

Answer: X = 19 or less than 19

2. Solve the expression by using addition property of order Y – 5 = 18

Answer: Y = 23

Types of Sampling Bias

Introduction :                

Sampling bias in static’s caused by some members of the population to be less likely to be included than others, where it results in a biased sample, It results in a biased sample, a non-random sample of a population in which all participants are not equally balanced or objectively represented If the bias makes estimation of population parameters impossible, the sample is a non-probability sample.

[Source Wikipedia].

Sampling Bias Types:

The types of sampling bias are :

Self modulate bias
Pre screening bias
Exclusion  bias
Over matched bias.
Self modulate bias:

The self modulated bias is used to deals with the group of certain people to know them about the previous and present calculations by the percentages using statics , This helps the people to know about the gain and loss percentage and the income difference between previous and past.

Example:

In 1988, the people spend tax as 1.25% for rupees 12000,but now they are paying 8% Means what is the percentage difference between Both ?

Solution:

For 12000 in 1988 1.25 % means = 150 rupees.

Now they are sending 8%

So, for 8% in 12000 means = 960 rupees.

The difference between the previous and present is 960 – 150 = 810 rupees.

For calculating the difference between the both self modulated biases is used.

Pre screening bias:

Pre screening bias is of trial participants, or advertising for volunteers within particular groups, this sampling is used to give the advantages and the disadvantages of the specific events about their weights loss sessions. 
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Bias Types :

Exclusion bias:

It occurred by the particular group of sampling present in the bias, for example the exclusion of subjects who have recently migrated into the study area , Excluding subjects who move out of the study area during follow-up is rather equivalent of dropout or nonresponsive, a selection bias in that it rather affects the internal validity of the study.

Over matched bias:

The over matched bias in sampling is used to tell about the features about the exposure terms present in radical terms, The over matched group becomes more similar to the cases in regard to exposure than the general population.