Breakingwave

Equations of Motion

2009-01-22T07:16:00.001-08:00

The 4 equations of motion deal with an object which is travelling with constant acceleration (which can be 0 and therefore constant speed).

The equations are as follows:

(1) v = u + at
(2) s= t(v+u)/2
(3) v²=u²+2as
(4) s = ut + at²/2

Where
a = acceleration – this must be constant for equations to hold
u = initial velocity, ie at the start of the journey
v = final velocity, ie at the end of the journey
s = displacement which is a vector quantity for the distance of the object from its starting point
t = time taken for journey
Any of these values can be found using the equations if at least 3 of the other values are known.

Equation (1) comes from the definition of acceleration as acceleration is the rate of change of velocity and therefore for constant acceleration
a = (v-u)/t which you can rearrange to make
v = u + at

Equation 2 can be found by either considering a distance time graph or using the average speed. (v+u)/2 gives the average speed during the journey as a is constant and buy multiplying this by t we find the displacement. Or from the graph we can find it as the area under the graph is a trapezium of height t and sides u and v so using the area of a trapezium formulae we find 2.

Equation (4) can be formed by substituting 1 into 2 so
s = t(u+at+u)/2
s = t(2u + at)/2
s= 2ut/2 + at²/2
s = ut + at²/2

Equations 3 can also be found using 1 and 2 by rearranging 2 to get an expression for t we find
t=2s/(u+v)

If we substitute this into v = u+at we get
v=u+2as/(v+u)
And by bringing up the v+u we find
v²+vu = u²+vu + 2as
And because we have vu on both sides we can cancel these to find 3

v²=u²+2as

Transformations of Graphs

2009-01-11T09:48:00.001-08:00

This looks at how a given graph will change when the the function is changed slightly eg how the graph y=x will change when it becomes y = 2x.

y = af(x)

The graph f(x) will "steeper" as the y value of each point is multiplied by a. It will appear like a "stretched" version of the graph y=f(x)

y =f(x) + a

The graph f(x) will move up by the amount a as a is added to each y value. This means that the points of intersection of the graph and the y axis will increase by the amount a. The intersection of the graph and the x-axis will depend upon the function of the graph.

y = f(ax)

This will make the graph appear "narrower" beacuse y is taking the value of f(x) a time across. So if the value of f(x) is 3 when x =12 then the value of f(4x) = 3 when x=3 as 12/4=3.

y=f(x+a)

This will shift the graph to the left by a. This is because the value of f(x) at x+a is displayed at the point x so effectively the graph occurs a earlier and therefore shifts to the left.

Sec, Cosec, Cot

2009-01-10T05:00:00.001-08:00

Sec, cosec and cot are all functions in trigonometry. They are simply equal to one over on of the other functions, ie cos, sin and tan.

Sec = 1/cos

Cosec = 1/sin

cot = 1/tan

You can remember which is paired with which using the third letter rule. This is that the third letter is the first letter of the corresponding function ie)

sec goes with cos
cosec goes with sin
cot goes with tan

Differentiate Inverse Sine (arcsin) - proof

2009-01-08T03:36:00.000-08:00

How to differentiate cos^-1x

y=cos^-1x

Bring the cos across

cosy = x

Differentiate both sides, remember when differentiating y time by dy/dx

-sin(y) dy/dx = 1

dy/dx = -1/siny

However we want to get the differential in terms of x, to do this we can use the identity

sin²t+cos²t = 1

so

sint = √(1 - cos²t)

putting this into our expression for dy/dx we get

dy/dx = 1/√(1-cos²y)

but cosy = x so

dy/dx = 1/√(1-x²)

by David Woodford

Differentiate Inverse Sine

2009-01-06T10:31:00.000-08:00

This tutorial explain how to differentiate inverse sine, this applies when using radians.

begin with

y = sin^-1 x
bring sin^-1 across to become sin
sin y = x
differentiate
cos y dy/dx = 1
note that the derivative of sint wrtt is cos t as explained in an earlier tutorial and by the chain rule when we differentiated sin y it became cosy time dy/dx as we are differnetiatiny a y and the derivative of y is dy/dx

then make dy/dx the subject

dy/dx = 1/cosy

We know the identity
sin²t + cos²t = 1
so we can wrtie
cos t =√(1 - sin²t)

we can now put this into the expression for dy/dx to get
dy/dx = 1/√(1 - sin²y)
but we know from the second line that sin y = x so

dy/dx = 1/√(1 - x²)

Factorising Quadratics

2009-01-02T06:10:00.000-08:00

Factorising quadratics is basically putting them in brackets. In this section we will look at two different ways of factorising quadratics( for simple and complex ones) and when they should be used.

Note/ sometimes a quadratic cannot be factorized using whole numbers, this is when you must use the quadratic equation to find the values of x. See my earlier post and c++ program

Simple type

Use when there is no coefficent of x²

eg)x²+2x-8

start by opening 2 brackets with an x in each

(x )(x )

put the first sign in the first bracket. If the second sign is + put the same sign in both, if its - put the opposite sign in the second bracket

(x+ )(x- )

find the 2 numbers that will add(if both signs in brackets are +) or subtract(if the signs in the brackets are different) to make the middle number(2) and multiply to make the end number(8)

(x+4)(x-2)

and thats your quadratic factorized

Complex Type

Use when there is a coefficent of x²

eg) 8x²-14x-15

Before we can open brackets we need to split up the 14x

8x² ?x ?x-15

the rule for the signs is the same as in the simple case, put the first sign before the first x term. If the second sign is + put the same sign in both, if its - put the opposite sign before the second x term

8x²-?x+?x-15

we also use the same rule for the to coefficients of x, they must add or subtract to make the middle number(14) but they must times to make the end number times the first(15×8=120)

8x²-20x+6x-15

we then take out the common factor of the first 2 terms(4x)

4x(2x-5) + 6x -15

we use the bracket(2x-5) as the common factor for the second 2 terms and find what we need to multiply by(3)

4x(2x-5)+3(2x-5)

we then take the 2 numbers in front of the brackets(4x and 3) as our second bracket

(4x+3)(2x-5)

and there we have a fully factorized quadratic

Matrix Multiplication

2009-01-02T06:09:00.000-08:00

This is a simple tutorial on how to multiply 2 matrices together. You can speed up doing this using my c++ matrix calculator (download matrix calculator) but I strongly recommend learning how to do them long hand first as you should only use formulae and program that you understand how they work. Multiplying matrices is a useful thing to know as it enables complex algebra to be simplified and i used extensively in 3d computer graphics.

What Matrices can i Multiply Together

You may fin it surprising that only certain size matrices can be multiplied together, this is because of the way the multiplication is done and how the size of the answer is determined.

The size of a matrix is given as rows by columns

eg) 2 3 1

3 6 1

is a 2×3 matrix. When multiply matrices the columns of the first matrix must be equal to the rows of the second. The size of the new matrix is given by the rows of the first by the columns of the second

eg) Fig1

How is the multiplication Done

The way in which matrix multiplication is done is surprisingly simple. Go to the first point in the new matrix. Look at the row it is in and go to the start of that row in the first matrix and the start of the column it is in in the second matrix. Multiply these to numbers together, then move along one in the first matrix and down one in the second matrix and multiply these together. Keep going till you have reached the end of the row of the first matrix, now add all of these values up and that is the first value for your new matrix.

So you go to the row in the first, the column in the second, times each pair of values and add them up for each value.

In the above diagram (fig 1)

u=ag + bk

v=ah+bl

w=ai+bm

x=cg+dk

y=ch+dl

and so on….

That may sound a little confusing but you’ll soon get used to it and it is very useful. If you want to practise and check your answers please download my c++ matrix calculator, the command for multiply is mlt, though you will have to define you matrices using the dim command first. All the data it needs off you is asked for

If you have any comments please leave them below or email me at woodford_4@hotmail.co.uk

Differentiating Trigonometry

2009-01-02T06:08:00.000-08:00

Sine and Cos Graphs Differentiating sin and cos

June 23, 2008

This is the basics of the sine cos and tan graphs and how sine and cos relate to give you tan. It also shows how to differentiate sin and cos.

The output or range of both sine and cos is from -1 to 1 when given any angle. They can be shown on a graph where y = sin(x) and y = cos(x). In these graphs all the angles go along the x axis and you can see a wave type shape is formed

Sine Graph

Cosine Graph

As you can see both the sin and cos graphs move periodically between -1 and 1 as the angles change, this pattern continues indefinitely because once you pass 360 degrees or 2 pi radians you will return back to the beginning. If you try to perform sin^-1 of a value out side the range -1 to 1 you will get an error.

Differentiate Sin and Cos

also notice that the gradient of the sin graph is the value of the cos graph for the same angle and that the gradient of the cos graph is the -value of the sin graph for that angle. This means that we can differentiate the sin and cos graphs:

if f(x) = sin(x) then f ‘ (x)=cos(x)

and

if f(x) = cos(x) then f ‘ (x) = -sin(x)

however if we use ax instead of x we must differentiate it by bringing the a out, when its just x this doesn’t matter as the differential of x is 1.

ie)

let y = sin(f(x))

now let u = f(x)

du/dx = f ‘ (x)

also

y=sin(u) as u = f(x)

dy/du = cos(u)

from the chain rule

dy/dx = du/dx * dy/du

therefore

if y = sin(f(x))

dy/dx = f ‘ (x)cos(f(x))

and similarly for cos

if y = cos(f(x))

dy/dx = -f ‘ (x)sin(f(x))

How to create a Facebook application

2009-01-02T06:07:00.000-08:00

This is aimed at people how already know programming and want to apply some of these skills to making a facebook application. You dont need to know much, even html will do though you wont make much of an application with this. Ive made mine using php and it was a site i had already developed.

Facebook is a great place to put your web apps and is easy to do but the documentation I could find on the site wasn’t that great (though mayb im just blind). First of all you need a facebook account, you should be able to do this easily enough, you also need an application, the application is no different to any other website and can be written in any language your familiar with (we just put it in a frame later on), i wrote my app when learning php with no intention of ever using it in facebook.

The first thing you need to do is get the developer application on facebook. Then choose to create an application by clicking set up an application. Choose a name and then click optional fields(this is where we set it up).

Callback URL is where you enter the address of your app, so if you have created a website you put a link to that sites home page. The canvas URL is the url for app on facebook, you have to choose a unique address for it which people can use to get to the app. If you just want to place your app in the page like i did, meaning that it works like a normall webpage, choose iframe under the canvas url box. We then choose application type to be website.

The next important field is “Can you application by added on facebook” and since your making an app for face book you probably want to set this to yes. You then need to fill in the install option page(if it can be added on face book). Here you want users to be able to add it to all pages. The default FMBL is the code that will appear on their facebook page when they add it. I set as a link to the apps home page.

The final section to fill in is the integration points that helps facebook work with your app. The only part you need to fill in here the side nav URL which is the link that will appear in the applications part of the users page when they use facebook, so they can get to your app.

Finally click save, after that you can mess around on the developer trying out the different parts, try adding an image in the settings part. You also have to develop some code so that you application actually does something have fun,

David Woodford

Complex and Imaginary Numbers

2009-01-02T06:06:00.001-08:00

Most of the time in maths you are working with “real” numbers ie) all the rational numbers so we have 1,3,3.5567,4/7,root 2 etc, however sometimes we will need to extend our field of numbers to imaginary numbers. This is when we want to find the root of a negative number eg root -9, as (-3)² and 3² both equal 9 we are unable to find the square root of -9. In order to solve this number we can use the “imaginary” number i such that

i = √-1

We then say that √-9 = 3i since √ab = √a √b and therefore √-9=√9 √-1 = 3i . This allows us to solve all equations with negative roots. It is also useful to know what iⁿ gives as shown below

i² = -1

i³=-i

i⁴=1

i⁵=i

Complex numbers are made up of both real and imaginary parts,

eg z =a+ib

is a complex number. When comparing complex numbers we need only compare the real and imaginary parts, by equation them to each other,

ie) if the complex number z=a+ib

and 2z = 6+10i

a=3 and b=5

To solve complex equations with fractions we often have to rationalise the fractions, like we would with surd’s, so that there are no imaginary parts on the bottom and we can compare the coefficient’s, this is done by multiplying by the conjugate.

note// the conjugate of z=a+ib is z’ = a-ib

eg) (2+3i)/(4+5i) = (2+3i)(4-5i)/(4+5i)(4-5i) = (23+2i)/41

Complex numbers can be represented on an argand diagram when the y axis is the imaginary part and the x axis is the real part, so that the number a+ib is plotted at (a,b), and a line is draw between this and the origin to represent the number. The length of the line is called the moduls and the angle between the real axis and the line is the argument.

Proof of Pythagoras Theorem

2009-01-02T06:03:00.000-08:00

Pythagoras theorem, that the square of the longest side of a right angled triangle is equal to the sum of the squares of the other 2 sides.

This can be proved quite easily by drawing a square into which fit 4 of the same right angled triangle as shown below

As you can see the area of the whole square is equal to the the sum of the 2 shorter sides squared or (a+b)². The area of the green square left is the square of the longest side c². We also know that the area of each of the triangles is 1/2 x base x height = ab/2

From these 3 areas we can prove the theorem. The know that the total area of the square is equal to the area of the green square plus 4 of the triangles ie)

(a+b)²=c²+ 4ab/2

a² + b² + 2ab = c²+ 2ab

The 2ab ’s cancel and we are left with Pythagoras theorem

a² + b² = c²

View the rest of my posts on this blog or at my website

Trigonometry Identities

2008-12-23T03:22:00.001-08:00

There a number of “identities” in trigonometry that can be found from the basic ideas of sin, cos and tan as explained in my earlier post. These identities can help in solving equations involving trig functions, especially when there are 2 or more different functions as the often allow you to write the equation in terms of one function, eg sin, that you can then solve.

One of the identities is:

sin² + cos² = 1.

To prove this consider a right angled triangle with side a,b and c as shown below

From this we can use Pythagoras theorem to say:

a²+b²=c²

now we know

sin t = b/c so b = csin t

cos t = a/c so a = ccos t

substituting these values in the above equation we get

(csint)² +(ccost)² = c²

canceling the c² we get

sint² + cost² = 1

There are trig functions that are equal to 1 over sin, cos and tan called cosec = 1/sin, sec = 1/cos and cot = 1/tan. These can be remembered using the third letter rule as the third letter of each of these corresponds to the the function it is one over.

Using these a cos² + sin² = 1 we can calculate other identities

tan²t + 1 = sec^{2^t}

We can obtain this by dividing through by cos² as we know sin/cos = tan, cos/cos = 1 and 1/cos = sec.

Other similar identities can be obtained for cosec and cot.

Economic Update 17th December

2008-12-17T08:46:00.000-08:00

A lot has happened again in the economic world this week including :

· Today the administrators that have taken over Woolworths said that all stores will close by January 5th.

· Unemployment has risen to 1.86 million.

· The pound hit a low of 1.0796 euros.

· Opec is making an output cut of 2.2million barrels per day.

· Yesterday the US interest rates to between 0 and 0.25%.

· National express is to cut 750 jobs

But on the positive side inflation has fallen to 4.1% !!!

Making Sense of the Economic Climate

2008-12-11T08:18:00.000-08:00

Yesterday the Bank of England cut interest rates to 2% (lowest level since 1951) but is it actually going to help us.
It will help those who are struggling to pay our mortgages as repayments should become cheaper but not all banks are going to pass on the cut/ the cut in fall to customers. They are much quicker to pass on the cut to savers however.
Those without a mortgage and have savings are losing out, they did not borrow money they could not pay back yet they are the ones who are getting less income from their savings.
Earlier in the week VAT was cut by 2.5% was this the right decision? In my opinion no, this is not going to get people to start spending money again as people have no money. People are likely to buy the same and keep the savings to pay for other outgoings .
The retail sector is really starting to feel the pinch Woolworths, MFI and The Pier have all gone into administration and we could see many more go in the next few weeks/months.
I don’t think that the economic situation is going to get any better in the near future infact i think that it is going to get wose but hopefully we have all learnt a valuable lesson, do not borrow so much money! and let us not go back to our borrowing ways
Nicola crowston

Hyperbolic Functions

2008-12-10T07:00:00.000-08:00

Hyperbolic functions are similar to sin,cos tan etc in trigonometry and share many similar rules. Usually hyperbolic functions are written like the trigonometric ones but with a h on the end, eg sinh and cosh.

The hyperbolic functions can be all written in terms of e, sinh and cosh are as follows

sinh(x) = (e^x - e^-x)/2
cosh(x) = (e^x + e^-x)/2

And tanh can be defined as sinh/cosh so

tanh = (e^x - e^-x) / (e^x + e^-x)

though this is often written as
tanh = (e^2x - 1) / (e^2x + 1)
by timesing the top and bottom by e^x

the other other hyperbolic functions sinh as sech, coth etc can be found in the same way as they would be in trigonometry, by using 1 over the other functions, ie sech = 1/cosh

Most of the identities in trigonometry have a similar identity with hyperbolic functions, however in most of these whenever there is a sin² it changes to a -sinh²
so
cosh² - sinh²=1
which you can work out by placing the equations with e’s in the place of sinh and cosh

3D Graphics

2008-12-05T08:37:00.000-08:00

This is a brief explanation of how to create basic 3D graphics. There are many different packages that will allow you to program using pre written 3D graphics library's, such as Java 3D but this is aimed at using and understanding the actual formula that can be used to draw the graphics yourself.

This assumes you are looking straight ahead from a single point and will map points in the 3D space in front of you onto a single plane at a fixed distance in front of you, probably 1 unit to keep it simple.

First of all consider the vertical direction, y axis. If we consider the right angled triangle formed by taking the line from you eye to the point, the line from your eye to beneath the point (ie the line where y =0) and the final verical equal to the y cordinate, we can use trigonometry to work out the height that need to be displayed.

From this we can work out the angle at your eye would be tan^-1(y/(x²+z²))
This is because the length of the "opposite" is equal to the y co-ordinate and the adjacent can be found using pythagoras from the x and z co-ordinates.

We now can use the angle we have found to work out where the point needs to be displayed in our plane. Again we need to form a triangle similar to the first one but where the top corner is the final position of the point when it is projected. Ie the opposite is the y co-ordinate we need.
therefore,
y = tan(t)((x/z)²+1)
y=y((x/z)²+1)
(x²+z²)

Matrix Calculator, C++ Program

2008-12-03T11:12:00.000-08:00

I recently went back to my c++ matrix calculator program and noticed it needed some improving so here are the improvements along with the source code.

Download the c++ matrix calculator here

Firstly I have added add/minus features to it. I accidentaly forgot to code these last time but have included them in the new program with the commands add and min. A note on the code, all the min function actually does is reverse all the signs on the second matrix and then call add.

I have also allowed you to determine both the number of rows and columns in the matrix to give it more flexibility. However this is not extended to the transformations (rotate and reflect) which still only work in 3 dimensions. Also in this process I have changed the lgth variable in the matrix to colls so that it now has rows and colls.

Finally to avoid some errors i have ensured all answer matrices are assigned colls and rows values.

Please feel free to use the program and don’t be put off by the fact it uses a command prompt, just type help to see the list of commands and when you type the command followed by enter it will give you step by step guidance on what to do.

David Woodford

// matrix.cpp : main project file.

#include “stdafx.h”

#include <iostream>

#include <cmath>

#include <string>

using namespace std;

class mat

{

public:

double mata[20][10];

int colls;

int rows;

void dimmat()

{

cout << “enter collums of matrix” << endl;

cin >> colls;

cout << “enter rows of matrix” << endl;

cin >> rows;

// cout << “” << endl;

for(int n=0;n<rows;n++)

{

cout << “enter values for row” << n << ” , each followed by ‘enter’:” << endl;

for(int m=0;m<colls;m++)

{

cin >> mata[n][m];

}

}

}

void display()

{

int rcount = 0;

int ccount = 0;

while(rcount < rows)

{

while(ccount < colls)

{

//cout << “matrix:”<<endl;

cout << mata[rcount][ccount] << ” , “;

ccount++;

}

cout << ‘\n’;

ccount=0;

rcount++;

}

}

/*void rotate()

{

mat mat2;

mat2.dimmat();

mat mat3;

mat3 = mult(this, mat2);

}*/

};

mat mult(mat A, mat B)

{

cout << “multiply started” << endl;

//char pause;

mat ans;

ans.colls=B.colls;

ans.rows=A.rows;

int rcount = 0;

int ccount = 0;

int c2 = 0;

while(rcount < 3)

{

//cout << “row: ” << rcount << ” started” << endl;

while(ccount < A.colls)

{

//cout << “collum: ” << ccount << “started” << endl;

ans.mata[rcount][ccount] = 0;

while(c2 < A.rows)

{

ans.mata[rcount][ccount] = ans.mata[rcount][ccount] + (A.mata[rcount][c2] * B.mata[c2][ccount]);

c2++;

}

//cout << “value is: ” << ans.mata[rcount][ccount] << endl;

//cin >> pause;

c2=0;

ccount++;

}

ccount=0;

rcount++;

}

if(A.colls != B.rows)

{

cout <<”matrices are wrong sizes to be multiplied” << endl;

return A;

}

else

{

ans.display();

return ans;

}

}

mat add(mat A, mat B)

{

mat ans;

if(A.colls == B.colls && A.rows == B.rows)

{

ans.colls = A.colls;

ans.rows=A.rows;

int i = 0;

int j = 0;

while(i < ans.colls)

{

j=0;

while(j<ans.rows)

{

ans.mata[i][j] = A.mata[i][j] + B.mata[i][j];

j++;

}

i++;

}

return ans;

}

else

{

cout << “matricies cannot be added : diffrent lengths” << endl;

return A;

}

}

mat minus(mat A, mat B)

{

int i =0;

int j =0;

while(i<B.colls)

{

while(j<B.rows)

{

B.mata[i][j] = 0 - B.mata[i][j];

j++;

}

j=0;

i++;

}

return add(A,B);

}

mat rotate(mat A, int angle)

{

mat T;

T.colls = 3;

mat matans;

//creat transformation matrix

double pi = 3.14159265;

double theta = (angle*pi)/180;

T.mata[0][0] = cos(theta);

T.mata[0][1] = 0 - sin(theta);

T.mata[0][2] = 0;

T.mata[1][0] = sin(theta);

T.mata[1][1] = cos(theta);

T.mata[1][2] = 0;

T.mata[2][0] = 0;

T.mata[2][1] = 0;

T.mata[2][2] = 1;

matans = mult(T, A);

return matans;

}

mat reflect(mat A, int angle)

{

mat matans;

mat T;

T.colls = 3;

//creat transformation matrix

double pi = 3.14159265;

double theta = (angle*pi)/180;

T.mata[0][0] = cos(2 * theta);

T.mata[0][1] = sin(2 * theta);

T.mata[0][2] = 0;

T.mata[1][0] = sin(2*theta);

T.mata[1][1] = 0 - cos(2*theta);

T.mata[1][2] = 0;

T.mata[2][0] = 0;

T.mata[2][1] = 0;

T.mata[2][2] = 1;

matans = mult(T, A);

return matans;

}

void input()

{

string dim = “dim”;

string com;

int end = 0;

mat matans;

int matcount = 0;

mat mats[11];

while(end == 0)

{

cout << “enter command>”;

getline(cin, com);

if(dim == com)

{

cout << “matrix” << matcount <<endl;

mats[matcount].dimmat();

matcount++;

}

if(com == “rot”)

{

cout << “which matirx?” <<endl;

int matnum;

cin >> matnum;

cout << “what angle (degrees)” << endl;

int rotang;

cin >> rotang;

matans = rotate(mats[matnum], rotang);

mats[10] = matans;

}

if(com == “rlt”)

{

cout << “which matirx?” <<endl;

int matnum;

cin >> matnum;

cout << “what angle (degrees)” << endl;

int rotang;

cin >> rotang;

matans = reflect(mats[matnum], rotang);

mats[10] = matans;

}

if(com == “ans”)

{

matans.display();

}

if(com == “mlt”)

{

cout << “first matrix” << endl;

int mat1;

cin >> mat1;

cout << “second matirx” << endl;

int mat2;

cin >> mat2;

matans = mult(mats[mat1], mats[mat2]);

mats[10] = matans;

}

if(com == “add”)

{

cout << “enter first matrix” << endl;

int A;

int B;

cin >> A;

cout << “enter second matrix” << endl;

cin >> B;

matans = add(mats[A],mats[B]);

mats[10] = matans;

matans.display();

}

if(com == “min”)

{

cout << “enter first matrix” << endl;

int mata, matb;

cin >> mata;

cout << “enter second matrix” << endl;

cin >> matb;

matans = minus(mats[mata], mats[matb]);

mats[10] = matans;

matans.display();

}

if(com == “dsp”)

{

cout << “which matirx?” << endl;

int matdsp;

cin >> matdsp;

mats[matdsp].display();

}

if(com == “let”)

{

cout << “which matirx?” << endl;

int mat1;

cin >> mat1;

int mat2;

cout <<”eaqual to (10 is answer matrix)” << endl;

cin >> mat2;

mats[mat1] = mats[mat2];

}

if(com == “help”)

{

cout <<”Davids Woodfords matrix calculator” << endl;

cout << “takes the following commands” << endl;

cout <<” ‘dim’    :: allows you to dfine a matrix”<< endl;

cout <<” ‘rot’    :: roates a matrix through an agnle” << endl;

cout <<” ‘rlt’    :: reflects a matrix through the line y=tan(a) where a is given” << endl;

cout <<” ‘ans’    :: displays the answer to the last calculation” << endl;

cout <<” ‘mlt’    :: lets u multiply 2 matricies together” << endl;

cout <<” ‘dsp’    :: displays a matrix specified”<<endl;

cout <<” ‘let’    :: allows u to assign one matrix the value of another, eg answer”<<endl;

cout <<” ‘add’    :: lets u add 2 matricies together” << endl;

cout <<” ‘min’    :: lets u minus 2 matricies together” << endl;

cout<<”===================================================================”<<endl;

cout <<”matricies are sotred in an array of 10, with numerical values starting at 0″<<endl;

cout<<”matrix 10 is the answer matrix”<<endl;

cout<<”any parameters will be asked for wen needed”<<endl;

}

}

}

int main()

{

cout << “Welcome to David Woodfords Matrix calculator” << endl << endl;

cout<<”type ‘help’ for a list of commands” <<endl;

// mat mata;

// mata.dimmat();

/* mat matb;

matb.dimmat();

mat matans;

matans = mult(mata, matb);

// mata.display();

*/

//rotate(mata, 30);

input();

return 0;

}

Tan=Sin/Cos

2008-12-03T11:11:00.000-08:00

this site is now at www.breakingwave.co.nr

This is often useful when solving trig equations so i thought i’d include it

basically:

sin = opp/hyp

and

cos=adj/hyp

sin/cos = ^(opp/hyp)/_(adj/hyp)

so if we cancel the hyp’s we get

sin/cos = opp/adj

and since tan = opp/adj

tan = sin/cos

Sine and Cosine Rules

2008-12-03T11:10:00.000-08:00

This is the basics of the sine cos and tan graphs and how sine and cos relate to give you tan. It also shows how to differentiate sin and cos.

Sine Graph

Cosine Graph

from the chain rule

dy/dx = du/dx * dy/du

therefore

if y = sin(f(x))

dy/dx = f ‘ (x)cos(f(x))

and similarly for cos

if y = cos(f(x))

dy/dx = -f ‘ (x)sin(f(x))

Area and Circumference of a Circle: pi

2008-12-03T11:08:00.000-08:00

This is a basic guide to using pi to find the area and circumference of a circle using pi. And also explores why pi makes our formulea work.

area =πr²

circumference = 2πr or πd

where r = radius and d=diameter

Area

First lets look at the area of a circle, given by area =πr². This is simple enough to use, we multiply the radius by itself and then by pi.

Does this make sense?

Well r squared is at least going to be an area but it might be a bit small so we multipy by pi. However this doesnt explain much untill we consider what pi is, the easiest way i find to do this is as follows

If we imagine a square that the circle fits inside perfectly(so it touches all four sides like the one above) r squared would give us one quadrant, so the area of that square is 4 x r² . Of course the circle’s area is a bit smaller so we need to find the ratio between the areas of the square and circle. If we then times this value by four we have a magic constant to multiply r squared by to find the area of a circle (we times by four because we need the area of 4 quadrants and r squared gives us one).

Now this magic constant is pi (which makes sense being just over 3, meaning the area of the circle is just over 3/4 of the area of the square).

Circumference

The circumference of a circle is given by 2πr or πd. This seems simple, we just multiply the diameter (2r) by our magic constant pi.

Does this also make sense?

seeing as we only have one r this time so only one length it seems we are just finding a factor to increase the length by to make a different length(the circumference) which makes sense.

Again lets consider the square into which our circle fits perfectly, the perimeter of this square would be 4 time the length of one of the sides.

Now the length of the sides = the diameter so the perimeter is 4d.

Notice again that the value we are trying to find for the square is multiplied by 4, but for a circle were gonna need a ratio thats a bit smaller.

So we need to replace the 4, for a square, with another, smaller, number — it seems pi will do the job.

Conclusion

To me when i consider pi i don’t look at it as a magical fundamental constant, but more a magical fundamental constant multiplied by four, because when I consider how these formula work using pi this is how they seem to work.

So this new constant is really the ratio of

area of square to area of circle

perimeter of square to circumference of circle.

and it = pi/4 = 0.785398….

so if you have a value for a square and you want a similar value for the circle you just need to multiply it by this number and you’ll have your answer

I welcome comments, improvements or errors in this post. Please leave your comments below or email me at woodford_4@hotmail.co.uk

thanks

Ellipse’s: equation, foci, eccentricity

2008-12-03T11:03:00.000-08:00

The equation of ellipse is usually written in one of 2 ways, as a parametric equation or as a Cartesian equation are as follows

P(h +a cos t ,k+ b sint), Where P is a point of the ellipse

where h and k are the amount the center of the ellipse is offset from the origin in the x and y directions and a and b the points on the x and y axis where the ellipse cuts.

An ellipse has to focal points (+/- ae) and can be defined as the loci of all the points P such that the sum of the distances from P to each focal point is constant.

The eccentricity of an ellipse e defines how sharp the ends of it are. e is always < 1 and can be calculated from the equation b² = a²(1-e²).

The directrices of an ellipse are x = ±a/e. These are lines such that the ratio of the distance from every point P on the ellipse from the directic and foci is the constant e, the eccentricity of the ellipse.

For example at the point (a,0) on the ellipse the distance to the directirc is

a/e - a = a(1/e - 1) = a(1-e)/e

and the distance from the ellipse to the foci is

a - ae= a(1-e)

the ration of these is

a(1-e)

a(1-e)/e

which equals e

and for any point p,

the distance to the directric is

a/e - acost = a(1/e - cost) = a(1-ecost)/e

when squared becomes

a²cost(e²cost-2e)+a²

e²

the distance to the foci is

√{(acost-ae)² + b²sin²t}

which with substitution for b given above becomes

a²cost(e²cost-2e)+a²

hence the ratio of the square of these 2 distances e² so the ratio of the distances themselves is e

Post by David Woodford, also avalible at www.breakingwave.co.nr