Solve Differentiation by Outcome

Introduction to solve differentiation by outcome:
Differentiation is one of the methods mostly used in calculus problems. A derivative of the function is called as differentiation process. Differentiation is used to find how the function changes as its input changes. It is used to for mention the quality of the function. In differentiation exam, we use different formulas. Trigonometric functions also used in differentiation process. I like to share this Define Differentiation with you all through my article.

Example Problems for Solve Differentiation by Outcome

Solve differentiation by outcome example problem 1:

Find the first derivative value of the given function g(y) = 16y2 - 6y

Solution:

Given function is g(y) = 16y2 - 6y

Differentiate the given function with respect to y, we get

g'(y) = 32y - 6y

Answer:

The outcome of the given function is 32y - 6y

Solve differentiation by outcome example problem 2:

Differentiate the given function f(x) = (sinx + 12tanx)

Solution:

Given function is f(x) = (sinx + 12tanx)

Differentiate the given function with respect to y, we get

f'(x) = (cosx + 12sec2x)

Answer:

The outcome of the given function is (cosx + 12sec2x)

Solve differentiation by outcome example problem 3:

Differentiate the function y = (14x + 3) (x + 3)

Solution:

Given function is y = (14x + 3) (x + 3)

Using uv formula,

(d / dx) (uv) = u (dv /dx) + v (du / dx)

Here, u = (14x + 3), v = (x + 3)

(dy / dx) = (14x + 3) * (d / dx) (x + 3) + (x + 3) * (d / dx) (14x + 3)

= (14x + 3) * 1 + (x + 3) * 14

Multiply the values and then add, we get

= 14x + 3 + 14x + 42

= 28x + 45

Answer:

The outcome of the given function is 28x + 45

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P Practice Problems for Solve Differentiation by Outcome

Solve differentiation by outcome practice problem 1:

Find the first derivative value of the given function g(x) = 12x16 + 6x15 dx

Answer:

The outcome of the given function is 192x15 + 90x14

Solve differentiation by outcome practice problem 1:

Differentiate the function y = (78x) (x + 1)

Answer:

The outcome of the given function is 156x + 78

Formula for Z Score

Introduction to formula for Z score:

Z score is an important concept in probability.  Subtract the mean from the raw score then divided by the standard deviation is called Z score.  Standard score and Z value are also called as Z score.  In this content, we are going to illustrate about “formula for z score” with related problems. I like to share this Percentile Formula with you all through my article.

Formula for Z Score – Formula for Z Score

Formula of Z score is `Z = (x - mu)/sigma`

Mean `mu`

Standard deviation `sigma`

Example Problems – Formula for Z Score

Example problem 1 – Formula for Z score

Calculate the Z score, for the value of x is 75, mean 25 and the standard deviation 7.

Solution:

Given,

The value of x is 75

The value of mean `mu` is 25

The value of standard deviation `sigma` is 7

Already we have the formula for estimating Z score

Formula: `z = (x - mu)/sigma`

`z = (75 - 25)/7`

= `50/7`

= 7.14

The value of Z score is 7.14

Answer:  7.14

Example problem 2 – Formula for Z score

Calculate the Z score, for the value of x is 115, mean 73 and the standard deviation 20.

Solution:

Given,

The value of x is 115

The value of mean `mu` is 73

The value of standard deviation `sigma` is 20

Already, we know the formula for estimating Z score

Formula: `z = (x - mu)/sigma`

`z = (115 - 73)/20`

= `42/20`

= 2.1

The value of Z score is 2.1

Answer:  2.1


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Practicing Problems – Formula for Z Score

Practicing problem 1 – Formula for Z score

Calculate the Z score, for the value of x is 7, mean 6 and the standard deviation 6.

Answer:  0.17

Practicing problem 2 – Formula for Z score

Calculate the Z score, for the value of x is 16, mean 4 and the standard deviation 5.

Answer:  2.4

Geometric Mean Formula

The geometric mean formula is the process of finding the average for the given set of data. The geometric mean is calculated by multiplying the given n numbers and then finding the square root for the multiplied number as the nth root. This geometric mean is similar to the arithmetic mean. This article has the details about the formula for geometric mean.

Steps to Find Geometric Mean Formula

Step 1: First multiply the given n numbers.

Step 2: Then find the nth root for the multiplied numbers.

The formula used to find the geometric mean value is

`root(n)(x1, x2 . . . .xn)`

Where x1, x2 . . . .xn are the given n numbers. Please express your views of this topic Discrete Random Variable Definition by commenting on blog.

Examples on Geometric Mean Formula

Example 1 for the formula for geometric mean:

Find the geometric mean for the data 5, 6, 8, 12.

Solution:

The given data for finding the geometric mean value is 5, 6, 8, 12.

In this given data first multiply the given set of data

5 x 6 x 8 x 12 = 2880

Then find the fourth root for the multiplied number, because there are four numbers.

Geometric mean = `root(4) (2880)`

Geometric mean = 7.325

The geometric mean value for the 5, 6, 8, 12 is 7.325.

Example 2 for the formula for geometric mean:

Find the geometric mean for the data 10, 11, 12.

Solution:

The given data for finding the geometric mean value is 10, 11, 12.

In this given data first multiply the given set of data

10 x 11 x 12 = 1320

Then find the cube root for the multiplied number, because there are three numbers.

Geometric mean = `root(3) (1320)`

Geometric mean = 10.97

The geometric mean value for the 10, 11, 12 is 10.97.

Example 3 for the formula for geometric mean:

Find the geometric mean for the data 1, 3, 5, 7, 9.

Solution:

The given data for finding the geometric mean value is 1, 3, 5, 7, 9.

In this given data first multiply the given set of data

1 x 3 x 5 x 7 x 9 = 945

Then find the fifth root for the multiplied number, because there are five numbers.

Geometric mean = `root(5) (945)`

Geometric mean = 3.93

The geometric mean value for the 1, 3, 5, 7, 9 is 3.93.

Interval

Introduction :-

An interval is an associated portion of the valid line. If the endpoints a and b are limited and are integrated, the interval is called closed and is denote [a, b].If the endpoint is not including, the interval is call open and denote (a, b).Here math, an interval is a locate of real numbers with the belongings that any number that lies connecting two numbers in the set is also integrated in the set. I like to share this All Real Numbers in Interval Notation with you all through my article.

Notations for Intervals:-

•    If one endpoint is comprise but not the other, the interval is denote [a,b) or (a, b] and is called a half-closed an interval [a, a] is called a sink interval.

•    Valid intervals have fun a key role in the premise of integration, because they are the simplest sets whose size or quantify or extent is easy to define.

•    The concept of measure can then be extend to more complicate set of real number, leading to the Boral measure and eventually to the Levesque measure.

•    Intervals are central to interval arithmetic, a general mathematical computing technique that automatically provides guaranteed enclosures for arbitrary formulas, even in the presence of uncertainties, mathematical approximations, and arithmetic runoff.

Integer Intervals:-

•    When a and b are integers, the notation [a ... b], {a ... b}, or just a... b is every now and then used to indicate the interval of all integers connecting a and b, counting both.

•    This information is used in some programming languages for example; it is used to define the set of valid indices of a vector.

•    An integer interval that has a predetermined lower or upper endpoint always includes that endpoint.

•    Therefore, the omission of endpoints can be explicitly denoted by writing a ... b - 1, a + 1... b, or a + 1 ...b - 1, or. Alternate-bracket notations like [a .. b] or [a .. b] are rarely used for integer intervals.

Classification of Intervals:-

The intervals of real numbers can be classified into eleven different types, listed below; where a and b is real numbers, with a < b:

•    empty: [b,a] = (a,a) = [a,a) = (a,a] = {} = O/

•    degenerate: [a,a] = {a}

•    proper and bounded:

•    open:(a,b) = { x | a
•    closed:[a,b] = { x | a `<=` x `<=` b }

•    left-closed, right-open:[a,b) = { x | a `<=` x
•    left-open, right-closed:(a,b] = { x | a< x `<=` b}

•    unbounded at both ends:(`oo`,`oo` ) =`RR`

Subtracting Math Time Problems

Introduction to subtracting math time problems:

Subtracting math time problems is entirely different from our normal real numbers substraction. Subtracting math time problems helps to know about the time difference between start and end work.  Subtracting math time problems only related to the time and here we must know about minutes and seconds etc for making subtraction ease.I like to share this Subtraction with Regrouping Word Problems with you all through my article.

Subtracting Math Time Problem Example:

Subtracting math time problems example1:

Solution:

Now subtract the time 2:10 from 5:30

Step 1:

The given times  are 2:10 and 5:30

Step 2:

Subtraction of 2:10 from 5:30

5:30-2:10=3:20

The answer for the above subtraction time problem is 3:20

Subtracting math time problems example2:

Solution:

Now subtract the time 1:22 from 12:34

Step 1:

The given times are 1:22and12:34

Step 2:

Subtraction of  1:22 from 12:34

12:34-1:22=11:12

The answer for the above subtraction time  problem is 11:12

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Subtracting math time problems example3:

Solution:

Now subtract the time 7:22 from 12:45

Step 1:

The given times are 7:22 and 12:45

Step 2:

Subtraction of 7:22 from 12:45

12:45-7:22=5:23

The answer for the above subtraction time  problem is 5:23.

Subtracting math time problems example4:

Solution:

Now subtract the time 3:59 from 12:45

Step 1:

The given time are 12:45 and 3:59

Step 2:

Subtraction of 3:59 from 12:45

12:45-3:59=8:46

Here 45-59=-14 which is less then zero.

So add 60 minutes and substract 1 from hours.

The answer for the above subtraction time problem is 8:46

Subtracting math time problems example5:

Solution:

Now subtract the time 6:47 from 10:23

Step 1:

The given times are 6:47 and 10:23

Step 2:

Subtraction of  6:47 from 10:23

10:23-6:47=3:36

The answer for the above subtraction time  problem is 3:36

Few more Subtracting Math Time Problem Examples:

Subtracting math time problems example6:

Solution:

Now substract the times 1:15 from 8:30

8:30-1:15=7:15.

The answer for the above subtraction time  problem is 54661

Subtracting math time problems example7:

Solution:

Now subtract the time7:34 from 11:34

11:34-7:34=4:00

The answer for the above subtraction time problem is 4:00

Subtracting math time problems example8:

Solution:

Now subtract the numbers  7:22 from 11:12

11:12-7:22=3:50

Here 12-22=-10 which is less then zero.

So add 60 minutes and substract 1 from hours

The answer for the above subtraction  time  problem is 730.

Subtracting math time problems example9:

Solution:

Now subtract the numbers 6:56 from 8:36

8:36-6:56=1:40.

Here 36-56=-20 which is less then zero.

So add 60 minutes and substract 1 from hours

The answer for the above subtraction  time problem is 1:40

Opposite Integers

Introduction - opposite integers:

The math integer is called natural numbers, opposite integer is nothing but negative integer (-), including 0, 1, 2, 3 ... and negatives of the non-zero natural numbers (-1, -2, -3....). Integers have negative integers, positive integers; integers are viewed as subset of the real numbers, positive and negative integer’s are-3, -2, -1, 0, 1, 2, 3.... In this article we shall discuss some arithmetic operations using opposite integers.Having problem with Consecutive Integers keep reading my upcoming posts, i will try to help you.

Example Problems - Opposite Integers:

Example 1:

Adding the opposite numbers:` (-31) + (-8)`

Solution: The sum of two positive numbers will be a positive numbers.

` (-31) + (-8) = -39`

Solution is `-39.`

Example 2:

Adding the opposite numbers: `(-13) + (-14) =?`

Solution: The sum of two positive numbers will be a positive numbers.

`(-13) + (-14) = -27`

Solution is` -27.`

Opposite integers -Subtraction of negative numbers:

Example 1:

Subtracting the opposite numbers: `(-1) - (-19)`

Solution: while subtracting the two numbers, the result carries sign of larger number

` (-1) - (-19) = 18`

Solution is `18` .

Example 2: Subtracting the negative numbers:` (-23) - (-5)`

Solution: The sum of two negative numbers will be a negative number.

`(-23) - (-5) = -18`

Answer is` -18.`

Example 3: Get the value form the expanded form` -9xx 10000 - 5xx 1000 - 5 xx 10`

Solution:  `-9xx 10000 - 5 xx 1000 - 5xx 10`

` = -90000 - 5000 - 50`

`=> -95050.`

Opposite integers - multiplication problems:

Example 1:

Multiply the two opposite numbers:` -11 xx-7`

Solution:

Here `-11i` s multiplicand

`-7` is multiplier

Step1: Multiply the ones term,

Step 2: Multiply the second term (tens)

Final answer is `-77.`

Example 2:

Multiply the two opposite numbers: `-12 xx -9`

Solution:

Here` -12` is multiplicand

`-9` is multiplier

Answer is` -108`

Numbers using words - opposite integers:

Example problem 1:

Write a number using words: `95`

Solution:

`=95`

`5` is ones place

`9` is tens place.

`=90+5 = ` ninety + five

So that` 95` is become in a word, ninety five.

Example problem 2:

Write a number using words: `65`

Solution:

`=65`

`5` is ones place

`6` is tens place.

`= 60+ 5= ` sixty + five

So that `65` is become in a word, sixty-five.Is this topic how to subtract integers hard for you? Watch out for my coming posts.

Practice Problems- Opposite Integers:

Problem 1:

Sum of positive integers: `(-7) + (-7)`

Result: `-14`

Problem 2:

Sum of positive integers: `(-7) + (-1)`

Result: `-8`

Problem 3:

Sum of positive integers:` (-7) + (-2)`

Result: `-9`

Probability of Prime Numbers

Introduction :

The probability can be explained briefly as the “N” number of possible out comes of the event, in that “n” number of possible put comes is favor for X. Then the probability of X is the ratio of number out comes favor for the X to the total number of possible event. This is called as probability. In this article we see about the probability of  prime numbers.

Formula Used for the Probability:

Statement: Probability is nothing but the ratio of possible out come for selected event to the total number of out comes.

Formula – To find the probability as



Let us see the problem using the probability for prime numbers.

Example Problem - Probability of Prime Numbers:

Example Problem - Probability of Prime Numbers:

Example 1:

In class room there are 20 members, First ten members are selected as A team and next ten members are selected as B team. How many of the prime numbers in A team only.

Solution:

To find the prime number in first ten members only

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, in that 1, 2, 3, 5, 7 are the prime numbers.

Thus

The total number of possible out comes = 10
Selected event as prime number in A team = 5
Formula: p(A) = `(n(a))/(n(s))`

P(A) = `(5)/(10)` = `1/2`

Answer: The probability of prime number in A team is ½


Example 2:

In a box consists of numbers, In that bow we pick numbers as follow.

3, 4, 8, 56, 23, 89, 11, 43, 65, 67, 9, 10, 15, 24, 2, 14, 13, 7, 29, 81

Solution:

Step 1: Find the prime numbers from the picked number in box.

That is 3, 23, 11, 67, 2, 13, 7, 29

Thus

The total number of possible out comes = 20
Selected event as prime number in picked number from box = 8
Formula: p(A) = `(n(a))/(n(s))`

P(A) = `(8)/(20)` = `(2)/(5)`

Answer: The probability of prime number in in picked number from box is  `(2)/(5)`

Solve Reflex Angle Problems

Introduction to reflex angles tutoring:

Reflex angles are special kind of angles in geometry.  Reflex angle is an angle which measure larger than 180 degree and smaller than 360 degree.  That is it lies between the two values180 degree and 360 degree. In this article we are going to explain about how to solve reference angle problems?

Brief Summary about Reflex Angles

Definition of Reflex angle:

Reflex angle is the angle between the two degrees 180 to 360.

Pictorial representation of a reflex angle:



Here ABC is a reflex angle.  Its measurement is equal to ∠ ABC= 315 degree it also represented as ∠ B = 315 degree.  Here B is called as the edge of an angle.  That is also known as vertex and then the rays AB and BC are represented as the arms of an angle.Please express your views of this topic 4th grade math problems and answers by commenting on blog.

Problems on Solve Reflex Angles

Problem 1:

Identify the following reflex angle and solve which angle is the reflex angle

a)       43 degree

b)       165 degree

c)       220 degree

d)       54 degree

Solution:

The answer is option C. 220 degree is a reflex angle. Because it is lies between the two degrees 180 to 360 degree.  So this is a reflex angle from the pair of angles.

Problem 2:

Identify the following reflex angle and solve which angle is the reflex angle

a)       171 degree

b)       310 degree

c)       55 degree

d)       101 degree

Solution:

The answer is option B. 310 degree is a reflex angle. Because it is lies between the two degrees 180 to 360 degree.  So this is a reflex angle from the pair of angles.

Problem 3:

Identify the following reflex angle and solve which angle is the reflex angle

a)       288 degree

b)       53degree

c)       169 degree

d)       16 degree

Solution:

The answer is option A. 288 degree is a reflex angle. Because it is lies between the two degrees 180 to 360 degree.  So this is a reflex angle from the pair of angles.

Intersecting Parallel Lines

Introduction:
The distance between the two lines are equal and they will never intersect, such lines are said to parallel lines. When a line cuts two parallel lines, it will form eight type of  angles. For intersecting parallel lines , we can also find the point of intersection distance using distance formula.

Angles of the parallel intersecting lines:

When the two parallel lines are cut by a intersecting line, then eight angles are formed. The name of angles are following

1. Corresponding angles

2. Alternate angle

• alternate interior angle
• alternate exterior angle

3.Adjacent angles

Angles in the Intersecting Parallel Lines:

Angles in intersecting parallel lines:

This diagram show intersecting parallel lines.



Angle 1 = angle 3 =angle 5=angle 7

Angle 2= angle 4 = angle 6= angle 8.

Adjacent angles:When parallel lines is cut by an intersecting line, then the adjacent angles sum up to 180 degree

Angle 1 +angle 2= 180 degree

Angle 3 +angle 4= 180 degree

Angle 5 +angle 6= 180 degree

Angle 7 +angle 8= 180 degree

Here the adjacent angles always up to 180 degree.

Corresponding angles:The angle in the same location around the intersection of two points is said to be corresponding angle.

When parallel lines are cut by an intersecting line, then corresponding angles are equal

Corresponding angles:Angle 1 = angle 6

Angle 2 = angle 5

Angle 3 = angle 8

Angle 4 =angle 7

Alternate angles:When parallel lines is cut by an intersecting line, then the alternate angles are equal

So Alternate interior angle:Angle 3 = angle 5

Angle 4 = angle 6

Alternate exterior angleAngle 2 = angle 8

Angle 1 = angle 7

Examples for Intersecting Parallel Lines:

Ex 1: Find the angle 2, when the angle 1 = 100°? Use the above diagram.

Solution:Here angle 1 and 2 are adjacent angles

When parallel lines is cut by an intersecting line, then the adjacent angles sum up to 180 degree

So angle 1+angle 2 = 180

100 +angle 2= 180

Angle 2= 180-100

Angle 2 =80 degrees

Ex 2: Find the angle 4, when the angle 1 = 110°?  Use the above diagram.

Solution:Here angle 3 and 4 are adjacent angles

Angle 1= angle 3 (By vertical opposite angles are equal)

Angle 3 = 110

When parallel lines is cut by an intersecting line, then the adjacent angles sum up to 180 degree

So angle 3+ angles 4 = 180

110 +angle 4= 180

Angle 4= 180-110

Angle 4 =70 degrees.

To finding the distance between two intersecting points of parallel lines.

Ex 3: Find the distance of line intersecting two parallel lines A(1,1) B(2,2)

Solution:Formula = v(x2-x1)2 + (y2-y1)2

= v(2-1)2 + (2-1)2

= v(1) + (1)

= v2 units

Right Angle Symbol

Introduction to right angle symbol:

Right angles are one of the generally frequent and interesting angles within every part of geometry. It even contains its individual angle symbol. Within the diagram below, locate the angle toward 90° and also observe to in its place of a small curve, the angle is noticeable through a tiny square symbol.  If you observe this, the angle determine within degrees if frequently not there.

Right Angle Symbol

A right angle is a ninety degree angle. The two lines to appearance a ninety degree angle are vertical to all other. The angle sign is within the appearance of a box. That sign specify to through a right angle. The angle mainly repeatedly utilize within the trade is the ninety degree angle. The two straight lines cross, 4 angles are formed. Those four angles for all time include up to 360°.

The main lines of the extent are two directly lines which pass through all other on 90° angles. Mainly plants also industrial unit are believed to be make down the main lines of the area. When lines pass through, the just time every four of the angles formed are equivalent toward all other is while all angle is a 90° angle. The ninety degree angle is called a right angle.Between, if you have problem on these topics formula for binomial probability, please browse expert math related websites for more help on normal distribution probability formula.

Example Problems for Right Angle Symbols

Example 1

In right angle triangle angle of triangle is 600, hypotenuse side of triangle is 30 apply the sine geometry to find the unknown side of the right angle triangle?



Solution:

Angle of triangle= 600 

Hypotenuse side of triangle = 30.

sin A =opposite/hypotenuse

sin 600 =`x/30`

sin 600 x 30= x

x = sin 600 x30

x =0.8660x30     {Known the value of sin 60 degree is 0.8660}

x=25.98

Opposite side = 25.98

Example 2

In right angle triangle angle of triangle is 500, opposite side of triangle is 12 apply the sine geometry to find the unknown side of the right angle triangle?



Solution:

Angle of triangle= 500 

Opposite side of triangle = 12.

sin A =opposite/hypotenuse

sin 500 =`12/x`

sin 500 x = 12

x = 12/sin 500

x =0.7660x12     {Known the value of sin 50 degree is 0.7660}

x=9.192

Hypotenuse side = 9.192

Review on Simplifying Radicals

Introduction to review on simplifying radicals:

Radicals are nothing but a root that also called as square root. Square root is mentioned with the symbol (sqrt) and sqrt (x). Normally, radicals are rewritten like, sqrt of (a) is rewrite as a (1/2).By the way radicals are expressed in various forms. The expression sqrt (8) is read as “radical eight”, or “the square root of eight”. Thus, we are going to see the review on simplifying radicals in different ways.

Formula for Review on Simplifying Radicals:

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

Multiplication property for radical expression: `n (sqrt (ab)) = n (sqrt (a)) n (sqrt (b))`

Division property for radical expression: `n (sqrt (a/b)) = (n (sqrt (a))) / (n (sqrt (b)))`

Different forms of radicals is,

A square (second) root is written as `(sqrt(a))` ,

A cube (third) root is written as `^3(sqrt(a))` ,

A fourth root is written as `^4(sqrt(a))` ,

A fifth root is written as:  `^5(sqrt(a))` .

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Example for Review on Simplifying Radicals:
Example for Review on simplifying radicals: `(sqrt (75))`

Given: `(sqrt (75))`

Solution: Given question says, radical (75),

When, we take root for 75 we obtain 25*3.

Because,                              `sqrt (1225)` = `sqrt (25) sqrt (3)`

= `5 xx sqrt (3)` 

Thus we can do Review on simplifying radicals in the prefered way.

Example for Review on simplifying radicals: `^3(sqrt (512))`

Given: `^3(sqrt (512))`

Solution: Given question says, cubic root of (512),

When, we take cubic root for 512 we obtain ( 8 * 8 * 8 )

Because,                              8*8*8 = 512

512 = `8^3`

Therefore, cubic root of (512) = `8^3`

= `8` .   

Thus we can do Review on simplifying radicals in the prefered way

Given: `(sqrt (81))`

Solution: Given question says, radical of (81),

When, we take square root for 81 we obtain (9 * 9).

Because,                           9*9 = 729

81 = `9^2`

Therefore, by simplifying the radical of (81) = `9`

Given: `^4(sqrt (1296))`

Solution: Given question says, cubic root of (1296),

When, we take fourth root for 1296 we obtain 6.

Because,                              6*6*6*6 = 1296

1296 = `6^4`

Therefore, by simplifying the fourth root of (1296) = `6`

Given: `^5(sqrt (16807))`

Solution: Given question says, fifth root of (16807),

When, we take fifth root for 3125 we obtain 8.

Because,                   7 * 7 * 7 * 7 * 7= 16807

16807 = `7^5`

Therefore, by simplifying  fifth root of (16807) = `7`

Unit Normal Vector

Vector function is different from scalar function because the vector function have the scalar value and direction normal vector is defined as the vector normal or perpendicular to the surface also unit vector is a vector whose length is 1 the unit normal vector is a vector perpendicular to the surface with a length value of 1 it is denoted by “hat” n.

Unit Normal Vector Formula

Below is the formula for unit normal vector:

The unit normal vector for the plane is` N = (dT)/(dphi)`

where dT= unit tangent vector

`dphi` = polar angle.

Unit tangent vector with ,

the normal vector

or

`N(t) =(T'(t))/(||T'(t)||)`

where `N(t)` = unit normal vector

`T(t)` = unit tangent vector = `(v(t))/(||v(t)||)`

Where v(t) = velocity vector = r’(t)

r(t) = differentiable vector

Solved Examples

Below is an example for unit normal vector:

Example: Find the unit normal vector for the vector valued function r(t) = t²i + t³ j .

Solution: Differentiable vector r(t) = t²i + t³ j

v(t) = r’(t)= `(d(t^2i + t^3j ))/(dt)`

v(t) = `2t i + 3 t^2j`

modulus of ¦v(t) ¦= `sqrt(4t^2+9t^4)`

= `sqrt(4t^2+9t^4)`

Therefore unit tangent vector T(t) = (v(t))/(¦v(t) ¦)

=  `(2t i + 3 t^2j)/(sqrt(4t^2+9t^4))`

T'(t) = `( [(2t i + 3 t^2j)(1/2) (8t+36t^3)(4t^2+9t^4)^(-1/2)-(4t^2+9t^4)^(1/2)(2i + 6t j)]) /(((4t^2+9t^4)^(1/2))^2)`

`= ((1/2)(16t^2i + 24t^3j+72t^4 i+108t^5j) (4t^2+9t^4)^(-1/2)-(4t^2+9t^4]^(1/2)(2i + 6t j))/ (4t^2+9t^4)`

=`(8t^2i + 12t^3j+36t^4 i+54t^5j) /((4t^2+9^4)^(1/2))-((4t^2+9t^4)^(1/2)(2i + 6t j))/ (4t^2+9t^4)`

=`((8t^2+36t^4)i + (12t^3+54t^5)j)/((4t^2+9t^4)^(3/2))-(2i + 6t j)/((4t^2+9t^4)^(1/2)) `

`=((8t^2+36t^4)i + (12t^3+54t^5)j)/((4t^2+9t^4)^(3/2))-((2 i+6t j)( 4t^2+9t^4))/((4t^2+9t^4)^(3/2)) `

`=((8t^2+36t^4)i + (12t^3+54t^5)j)/((4t^2+9t^4)^(3/2))-((8t^2+18t^4)i+(24t^3+54t^5)j)/((4t^2+9t^4)^(3/2)) `

=`((8t^2+36t^4-8t^2-18t^4)i + (12t^3+54t^5-24t^3+54t^5)j)/((4t^2+9t^4)^(3/2)) ` 

T’(t)= (18t^4 i + (-12t^3)j)/((4t^2+9t^4)^(3/2))

||T'(t)|| = `sqrt(((18 t^4)^2+(-12t^3)^2)/((4t^2+9t^4)^(3/2))) `

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Therefore the unit normal vector    

N(t) =(T(t))/(||T'(t)||)

By substituting T'(t) and ||T'(t)|| value in N(t) we get,

`= (((18t^4i + (-12t^3)j))/((4t^2+9t^4)^(3/2)))/(((18 t^4)^2 + (-12t^3)^2)/(sqrt((4t^2+9t^4)^3)))`

`=((18t^4i -12t^3j)((4t^2+9t^4)^(3/2)))/(((4t^2+9t^4)^(3/2))((324t^8-144t^6)^(1/2)))`

`=(18t^4i -12t^3j)/((324t^8-144t^6)^(1/2)) `

By simplifying above equation we get N(t)` = (9t i-6j)/ (sqrt(9t^2-4)) `

Therefore the unit normal vector for the vector valued function r(t) = t²i + t³ j is

N(t)=`(9ti-6j)/ (sqrt(9t^2-4))`

Learn Experimental Probability From Experts

The learning probability started with gambling, the probability mostly used in the field of physical science, commerce, biology science, medical science, weather forecasting, etc.,    an experimental having more than one possible outcome is called statistical experimental. Probability is a method which numerically measures the degree of uncertainty and, there, of certainty of the occurrences of an event. A statistical experimental is also known as a trial. In this article we shall learning an experimental probability.

Experimental Probability Examples

Example 1: A coin is tossed 50 times. 25 times head appear. Find the experimental probability of getting heads.

Solution:

Step 1: Experimental probability = number of times the event happen / total number of trials.

Step 2: Number of times heads appeared = 25.

Step 3: Total number of experiments = 50.

Step 4: Therefore experimental probability of getting a head = 25/50= 1/ 2

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Example 2: John plays game his college. He did not score in 29 out of 58 games he played. What is the experimental probability of john score in a game?

Solution:

Step 1: Number of game john make score = 58 – 29=29

Step 2: P (john make a score) = Number of games in which john make goals / Total number of games john play

Step 3:=29 / 58

= 1 / 2.

Step 4: Probability of john make a goal in a game is 1 / 2.

Example 3:

A bag contains black, green and white cards. Kim picked a card at random for 40 times. Find the experimental probability of pick up a green card.

Color Card picked number of times

Black   15

Green    5

White   20

Solution:

Step 1: P = Number of time card picked / Total number of trial.

Step 2: 5/40= 8.

Step 3: The experimental probability for green card is 8.

Practice Problems

Rock plays his school. He did not make score in 50 out of 100 games he played. What is the experimental probability of rock scoring in a game?

Answer: 1 / 2.

Rose plays his school. She did not make score in 30 out of 60 games he played. What is the experimental probability of rose scoring in a game?

Answer: 1 / 2.

Definition Permutation

Introduction :

Fundamental principle of counting states that “if an event can occur in m different ways, following which another event can occur in n different ways, then the total number of occurrence of the events in the given order is m x n.” Let us find the number of 4 letter words, with or without meaning that can be formed out of the letters from the word ROSE, where we could repeat the letters. Since we have to find 4 letter words, draw 4 squares `square square square square` . Keep in mind that repetition is not allowed. The first place can be filled in 4 different ways and the remaining three letters in three different ways, following which the third place can be filled in 2 different ways; following which, the fourth place can be filled in 1 way. Thus, the number of ways in which the 4 places can be filled, by the multiplication principle, is 4 x 3 x 2 x 1 = 24. Hence, the required number of words is 24.Here we can see teh definition of permutation.

Permutation and Definition of Permutation:

In the above example, we are actually counting the different possible arrangements of the letters such as ROSE, ROES, SOER,… etc. Here, in this list, each arrangement is different from the other. In other words, the order of writing the letters is important. Each arrangement is called a permutation of 4 different letters taken at a time. Now, if we have to determine the number of 3 letter words, with or without meaning, which can be formed out of the letters of the word NUMBER, where the repetition of the letters is not allowed, we need to count the arrangements NUM, NMU, MUN,…etc. Here, we are counting the permutations of 6 different letters taken 3 at a time. Required number of words = 6 x 5 x 4 = 120.


Definition of Permutation:

A permutation is an arrangement in a definite order of a number of objects taken some or all at a time.

Formula for nPr = `(n!)/((n-r)!), 0<=r<=n`

Compound Events

Introduction to compound events:

Numerical measure of the likelyhood of an event to occur. If in an experiment there are n possible ways exhaustive and  mutually exclusive and out og them in m ways int he A occurs . then the probability of occurrence of the event A is given by ,

P(A) = m / n

Compound Events:

An experiment consists of two or more simple events called as compound events.

Examples:

Rolling a die
Tossing two dice and two coins

Types of Compound Events :

There are two types in compound events ,

Independent compound Event
Dependent   compound Event


Independent Event :

In compound events, when the outcome of one event does not change the outcome of the second event, these are called independent events.

The probability of two independent events is obtained by multiplying the probability of the first event by the probability of the second event.

Formula:

The probability of compound events when they are  Independent Event is,

P (A and B) = P (A) . P (B)      [A and B are two events]

Dependent Event:

Dependent events are two or more events where the outcome of one does affect the outcome of the other.

If 2 events, u and v are mutually exclusive, then the probability that either U or V occurs is the sum of the probability of u and probability of v.

Formula :

P(u or v) = P(u) + P(v)
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Example Problems for Compound Events

Example compound events problem 1:

Jack is going to a stationary shop to buy stationary items. There are 20 ball point pens, 35 gel pens and 15 parker pens. Jack randomly chosen one pen.  What’s the probability that john picked one is ball point or gel?

Solution:

Total no .of pens = 20 + 35+15 = 70

Since a pen cannot be both ball point and a gel, the events are mutually exclusive. Hence,

P (ball point or gel) = (no. of ball point pens + no. of gel pens) / Total no. of pens

= (20 + 35) / 70

= 11/ 14.

Answer:

The final answer is (11 / 14)

Example compound events problem 2:

There are three black cats and four grey cats in a cage, and none of them want to be in there. The cage door opens briefly and two cats escape. What is the probability that both escaped cats are black?

Solution:

Total number of cats = 3 black cats + 4 grey cats = 7

Probability of first cat escaped is black = 3 / 7

P (next one also black) = 2 / 6

Probability of both escaped cats is black:

= 3 / 7 * 2 / 6

= 6 / 42 = 1 / 7

This is an example for dependent event .

Permutation and Combination Help

Permutation and Combination Help

Permutation is the process of rearranging the given number of elements or objects. Permutation avoids the repletion. Combination is a selection of things or objects or any other from a collection. In combination order of objects is irrelevant.

Permutation formula: P(n,r) = `(n!) / ((n-r)!)`

Combination formula: C(n,r) = `(n!) / (r!(n-r)!)` ,

where, n gives number of things, and r gives number of times.

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Permutation and Combination Help – Solving Example Problems

See these example problems, it will help you to understand about how to find permutations and combinations.

Example 1: In how many ways can 4 pens be arranged from among 12 pens?

Solution:

This problem involves 12 pens, chosen 4 at a time.

Permutation formula: P(n,r) = `(n!) / ((n-r)!)`

Here, n = 12, r = 4

P(12,4) = `(12!)/((12 - 4)!)` = `(12xx11xx10xx9xx8!) / (8!)` = 11880

There are 11880 possible ways to arrange 5 pens from among 12 pens.

Example 2: In a jar there are 12 balls. In how many ways can 4 balls be chosen from a jar?

Solution:

This problem involves 12 balls, chosen 4 at a time.

Combination formula: C(n,r) = `(n!) / (r!(n-r)!)` ,

Here n = 12, r = 4

C(12,4) = `(12!) / (4!(12 - 4)!)` = `(12xx11xx10xx9) / (4xx3xx2xx1)` = 495

Therefore in 495 different ways can 4 balls be chosen from 12 balls.

Example 3: Calculate the number of permutations and combinations; n = 11, r = 5.

Solution:

Permutation: P(n,r) = `(n!) / ((n-r)!)`

Combination: C(n,r) = `(n!) / (r!(n-r)!)`

Here, n = 11, r = 5.

P(11,5) = `(11!) / ((11 - 5)!)` = `(11xx10xx9xx8xx7xx6!) / (6!)` = 55440

C(11,5) = `(11!) / (5!(11 - 5)!)` = `(11xx10xx9xx8xx7) / (5xx4xx3xx2xx1)` = 462

Therefore permutations = 55440, combination = 462.

Permutation and Combination Help – Solving Practice Problems
Solve these following practice problems, it will help you to get practice on how to find the permutation and combination.

Problem 1: In how many ways can 2 books be arranged from among 10 books?

Problem 2:  In a jar there are 15 orange balls. In how many ways can 6 balls be chosen from a jar?

Problem 3: Calculate the number of permutations and combinations; n = 9, r = 4.

Answer: 1) 90 2) 5005 3) P = 3024, C = 126

Vector Space Properties

Introduction to vector space properties:

A vector space is a set V consists of two operations: Vector addition and scalar multiplication. These operations satisfy assured properties. The scalars are taken as of a field F; anywhere F stands any for the real numbers R otherwise the complex numbers C. Consider the set V over a field F, if specified an operation vector addition describe in V, indicated v + w for every v, w in V, and an operation scalar multiplication in V, indicated b * v for every v in V and b in F. Let us see about the vector space properties.
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Description - Vector Space Properties:

Consider the vector space V is a set then the two operations addition and multiplication that satisfy the following properties.

Consider the two elements u and v in vector space V, after that u + v is and element of V .

u + v  =  v + u

u + (v + w)  =  (u + v) + w

There is an element 0 in V such that

u + 0  =  0 + u  =  u

For every u in V there is an element -u with

u + (-u)  =  0

If u is in V and b is a real number after that b*u is in V

b * (u + v)  =  b * u + b * v

(b + c) * u  =  b * u + c * u

b * (c * u)  =  (bd) * u

1 * u  =  u

Vector spaces contain numerous additional properties.  Followings are the some of the most basic ones.

Consider V is a vector space subsequently

0 * u = 0 for all u in V

c * 0 = 0 for all scalars c

If cu = 0 then either c = 0 or u = 0

(-1) u = -u for all u in V

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Example Problems for Vector Space Properties:

Example 1 for vector space properties

If u is vector then prove (-1) u = -u

Proof:

(-1) u = -u

(-1) u + u = (-1) u + (1) u 

(-1 + 1) u = (-1) u + (1) u 

0 * u = (-1) u + u

(-1) u + u = 0

Thus, (-1) u = -u

Example 2 for vector space properties:

If a and b are vectors then prove a + (b - a) = b

Proof:

a + (b - a) = a + (b + (-1) a)

= 1a + ((-1) a + b)

= (1a + (-1) a) + b

= (1 + (-1)) a + b

= (0 * a) + b

= 0 + b

= b
Proved that a + (b - a) = b

My Previous Blog :- http://advancemath.blogspot.in/2012/10/how-to-solve-matrix-equality.html

Semi Circle Geometry

Introduction :
In geometry, a semicircle is half circle, it is created by the cutting a whole circle by the side of a diameter line. In semi circle, any diameter of a circle is cuts it into the two equivalent semi circles. In geometry, a semi circle is used to construct two lengths using straight-edge and compass. A triangle decorated in the semi circle is always s right triangle.

More about Semicircle:

In geometry, the perimeter of a semicircle is not partially the perimeter of a circle.
So the formula for finding a perimeter of semi circle is, ?R + 2R
The r is defines radius of the semi circle.
The p is Pi, the Pi value is approximately describe 3.142
The final formula for finding the perimeter of the semi circle is,
R (p+2)

Area of a semicircle:

In geometry, the area of a semicircle is half the area circle.
The Formula for finding the area of semicircle is,
In this area of semi circle formula, the R is defines radius of the semicircle.

? is defines exact value 3.142

Example of area of semi circle:
A semicircular shaped doormat has a radius of 12cm.
Find the area of the doormat.
The given R value is 12.


In this area of semi circle formula, the R is defines radius of the semicircle.
? is defines exact value 3.142
Example of area of semi circle:

A semicircular shaped doormat has a radius of 12cm.
Find the area of the doormat.
The given R value is 12.

= 113.04

So, the area of semicircle is 113.04 square cm.


Uses of Semi Circle:

In geometry, the semicircle is mainly used to construct the arithmetic and geometric.
It means the length of two straight edges and compass.
If we would like to generate a semi circle with a diameter of a + b.
The length of semicircle and its radius are half of the diameter.
The geometry is make a diameter into two segments of lengths a and b.
Then those segments are connecting to common with the semi circle perpendicular diameter.

Substitution Elimination Method

Introduction to substitution elimination method:

These methods are been used to calculate the unknowns involved in different equations. Mathematically speaking, we need at least n number of equations, if there involved n unknowns. Here, let us discuss the substitution elimination method to solve simultaneous equations of two unknowns.

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Example on Substitution Elimination Method:

Ex 1: Solve the following equations using elimination method:

x + 2y = 10

x – y = 4

Sol:  Here let us call the equations as 1 and 2.

x + 2y = 10 ---------`->` (1)

x – y = 4 ------------`|->` (2)

(2) × 2  `rArr` 2x – 2y = 8

(+) (1)    `rArr` x + 2y = 10

`rArr` 3x = 18

`rArr` x = `(18)/(3)` = 6

`rArr`   x = 6

Now plug in x = 6 in the equation number (1) as follows:

x + 2y = 10 `rArr` 6 + 2y = 10

`rArr` 2y = 10 - 6 = 4

`rArr` y = `(4)/(2)` = 2.

`rArr` y = 2

Therefore, the solution of the given equations are x = 6, y = 2.

Example on Substitution Elimination Method

Ex 2: Solve the following equations using substitution method

x + 2y = 10

x – y = 4

Sol:  Here let us call the equations as 1 and 2.

x + 2y = 10 ---------`|->` (1)

x – y = 4 ----------`|->` (2)

(2)  `rArr` y = x – 4 --`|->` (3)

Now, substitute y = x – 4 in (1), we get as follows:

x + 2y = 10     `rArr`  x + 2 (x – 4) = 10

`rArr` x + 2x – 8 = 10

`rArr` 3x = 10 + 8 = 18

`rArr` x = 18/3

`rArr` x = 6

Now plug in x = 6 in the equation number (3) as follows:

y = x – 4  `rArr` y = 6 – 4 = 2                                                     

`rArr` y = 2

Therefore, the solution of the given equations are x = 6, y = 2.

Practice Problems on Substitution Elimination Method

Solve the following equations using substitution elimination method:

1. 3x + 2y = 23

x – y = 1 

Answer: x = 5, y = 4.

2. 2x + y = 21

x + y = 15

Answer: x = 6, y = 9.

My Previous Blog :- http://advancemath.blogspot.in/2012/10/perimeter-semi-circle.html