Examples of Locus

Topic:- Locus

Locus refers to a collection of points.

It's generally used to define a figure

Example 1:


A circle is a locus of all points ,Which are equidistant from a fixed point ,called the center of the circle.


Example 2:

A line is the locus of all points equidistant from two fixed points or from two parallel lines.

Let us consider a line

C_____________________________D >

-   -   -   -   -   -  -
A_____________________________B >
 k  l  m  o  n  p  q

Here,AB is a collection of points k,l,m,n,o,p,q
So,AB is called as Locus of all these points.

Commom point:
All the points are equidistant from line CD.

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Using Shell or Disc Method to Find Volume of the Solid

Disc method and Shell(cylinder) method of integration are the two different methods of finding volume of solid of a revolution, using rectangular coordination system the functions are defined in terms of x in the below problem.

Topic : Disc or Cylinder Method of Finding Volume of the Sphere.

Problem : Use the disc or shell method to find the volume of the solid generated by revolving the regions bounded by the graphs of the equations about the x axis. y=x³ y=0, x=2

Solution :

y = x³ => ³√y = x
or (y)^1/3 = x
or x = y^1/3

Volume of a Solid by rotating about x-axis is given by:

V = 2π_a∫^bp(y)h(y) dy
here p(y)=y^1/3, h(y)=y
when x = 2 and y = 3³ = 8
So a = 0 and b = 8
Plugging in all the values in the formula, we get

V = 2π₀∫⁸(y)^1/3.y dy

= 2π₀∫⁸(y)^4/3 dy (as 1/3 + 1/1 = (1+3)/3 = 4/3)

= 2π[y^7/3/(7/3)₀]⁸ (as 4/3 + 1/1 = (4+3)/3 = 7/3)

= 2π[(8)^7/3/(7/3)- (0)^7/3/(7/3)]

= 2π[((2)³)^7/3/(7/3)]

= 2π(2⁷/(7/3))

= 2π(128/(7/3))

= 2 * 3 * 128 * π / 7

= 768π /7


So this how the volume of Solid of revolution is determined when the equations about the x axis.
For more help write to our calculus help.

Problem on Trigonometric Identity

Trigonometric Identities are the standard equations used to solve trigonometric problems.
Below is as one such problem.

Topic : Trigonometric Identity

A proof for Trigonometric Identity

Problem : Prove the given identity 2 Cos x - 2 Cos³ x = Sin x Sin 2x

Solution :

LHS : 2 Cos x - 2 Cos³ x

= 2 Cos x (1 - Cos²x)
= 2 Cos x . Sin² x
= 2 Cos x . Sin x . Sinx
= Sin x . 2 Sin x Cos x
= Sin x . Sin 2x (using Sin 2x = 2 Sinx Cox x)

So we proved 2 Cos x - 2 Cos³ x = Sin x Sin 2x

If you have any queries related to the proof please leave a comment and we will get back to you soon.

A Solved Problem on Simplification

Topic : Simplification

Problem : Simplify (c + 9)(c³ - 3c² - 7c)

Solution :
(c + 9)(c³ - 3c² - 7c)
= c^4 - 3c³ - 7c² + 9c³ - 27c² - 63c
= c^4 + 6c³ - 34c² - 63c

Solving quadratic equation

Problem on how to solve a quadratic equation:

Know more about quadratic equations http://en.wikipedia.org/wiki/Quadratic_equation

Question:
(x + 2) / (x - 3) = 42 (x + 7)

Solution:

(x +2 ) (x + 7) = 42 (x - 3)

x² + 7x + 2x + 14 = 42x - 126

x² + 9x + 14 = 42x - 126

x² + 9x + 14 - 42x + 126 = 0

x² - 33x + 140 = 0

Using Factorization

x² - 28x - 5x + 140 = 0

Taking out common factors

x(x - 28) -5(x - 28) = 0

(x - 5) (x - 28) = 0
x -5 = 0
OR
x - 28 = 0

Either

x = 5

OR

x = 28

For more help on solving quadratic equations.

Question to Find Summation and Draw Box Plot for all Quartiles

Topic : Box Plot

Question : For given amount 95000, 0 and - 65000 , Function P(Xi) 0.60, 0.20 and 0.20 respectively draw a neat Box Plot and represent quartiles.

Solution :

Question on Permutation of Zeros in factorial 500

Topic : Permutation
Question : How many zeros are at the end of factorial 500?

Solution :
Highest power of a prime p that divides n!
Let n be a natural number, and let p be a prime. Then the exponent of the highest power of p that divides n! is equal to
N = [n/p] + [n/p2] + [n/p3] + ..... where [x] denotes the greatest integer less than or equal to x.
The highest power of 2 which divides 500! is = [500/2] + [500/2^2] +[500/2^3] + [500/2^4] +[500/2^5] + [500/2^6] +[500/2^7] +[500/2^8] +[500/2^9]
= 250+125+62+31+15+7+3+1 = 494.
The highest power of 5 which divides 500! is = [500/5] + [500/5^2] +[500/5^3] + [500/5^4]=100+20+4 =124
hence the highest power of 10= 5*2 which divides 500! is 124
Hence the number of zeros in 500! is 124.

Another method to solve the same problem.
you get a zero at the end of a product if one of the factors contains a prime factor of 5 and another of the factors contains a prime factor of 2. The prime factors of 2 and 5 multiplied together make 10 and produce a 0 at the end of the product.

If you want to find the number of 0's at the end of 500! in base 10, you need to find the total number of factors of 5 and the total number of factors of 2 in all the numbers from 1 to 500; the smaller of those two numbers will be the number of 0's at the end of 500!

Here is how you would determine the number of 0's at the end of 500!

1) 1 out of every 5 numbers 1 to 500 inclusive (100 of them) contains at least one factor of 5.
2) Of the 100 numbers 1 to 500 inclusive that contain at least one factor of 5, 1 out of every 5 (20of them) contains a second factor of 5.
3) Of the 20 numbers 1 to 500 inclusive that contain a second factor of 5, 1 out of every 5 (4 of them) contains a third factor of 5.
The total number of factors of 5 contained in the numbers 1 to 500 inclusive is then
100 + 20 + 4 =124
in short :
500/5 = 100
100/5 = 20
20/5 = 4
4/5 = 0
----
124
( Write only the quotient)
similarly, The total number of factors of 2contained in the numbers 1 to 500 inclusive is then

500/2 = 250
250/2 = 125
125/2 = 62
62/2 = 31
31/2 = 15
15/2 = 7
7/2 = 3
3/2 = 1
1/2 = 0
-----
494
so there are 124 factors of 5 , 494 factors of 2 and we get 124 combinations of 5x2 ( match the 124 5's with the 124 2's -- no more 5's left -so we get only 124 combinations of 2 x5 ).thats why we take the smaller of the above two calculations.
(however, 500/2= 250 > 124 so we need not perform the other calculations.)
and the answer is 124 0's at the end of 500 !