Maths Coursework:- T-Total

(t + 3) + (t - g + 3) + (t - 2g +3) + ( ( t + 3 ) - ( 2g - 1 ) ) + ( ( t + 3 ) - ( 2g + 1 ) )

t + 3 + t - g + 3 + t - 2g + 3 + t + 3 - 2g + 1 + t + 3 - 2g - 1

5t - 7g + 15

To understand this information better I need to find out why the numbers in front of the terms are as they are. The 5t is present because there are 5 separate boxes within the T-shape which, when added produce the T-number. The -7g is present for one reason. When you move up from the t the line above is -g then the line above it is -2g. However on the top line there are three numbers in a horizontal row all at -2g. It is therefore -7g as:

3 ( - 2g ) + 1 ( - g ) = - 7g

The next step is to discover why there is the number 15 in the newly translated formula. We know that there are still 5 different squares within the T-shape and that the shape was translated 3 across the x-axis. If we multiply 3 by 5 we get 15 but I need a way of generalising this. Every square you move along the x-axis increases t by 1 so we need something to represent this. It shall be x. I will now see if placing this in the formula works.

+ 3 across

This is what happens when the numbers are placed into the new Translation formula. The formula Should give the answer of 55 as being the new translated T-total.

5t - 7g + 5x

( 5 x 22 ) - ( 7 x g ) + ( 5 x 3) = 110 - 70 + 15 = 55

As you can see the formula gave the correct answer but now I need to see what happens when the T-shape is translated up and down th y-axis.

I have translated the T-shape +3 down on the y-axis and the results are shown. Everytime the T-shape is moved down 1 on the y-axis g is added to t. If we now add the terms in the newly translated T-shape you get this:

( t + ( g - 1 ) ) + t + g + ( t + ( g + 1 ) ) + t + 2g + t + 3g = p

t + g - 1 + t + g + t + g + 1 + t + 2 g + t + 3g = p

5t + 8g = p

Looking at this we can see that also for every 1 down the y-axis the amount that g is multiplied by increases by 5. So that one down would be -2g, two down would be 3g and finally 8g for three down. Every square you move along the y-axis adds g to t so we need something to represent this. It shall be y. However we also need to account for the 5g increases. I shall see if including 5gy in the formula works. I will use a numerical example to prove the formula. If the formula works then the answer should be 190.

5t - 7g + 5gy = p

( 5 x 22 ) - ( 7 x 10 ) + 5( 10 x 3 ) = 190

The formula gives the correct answer but now I have two separate formulae for translation. I now need to combine

The two to find an overall formula to find the T-total when the T-shape is translated anywhere about the grid.

I'am now going to try a combined Translation. The formula I am going to try is a straight combination of the two separate formulae. If the formula is right then the answer will be 210.

5t - 7g + 5x + 5gy = p

( 5 x 22 ) - ( 7 x 10 ) + ( 5 x 4 ) + 5( 10 x 3 ) = 210

The answer was correct. Therefore the formula for finding the T-total when the T-shape is moved anywhere around the grid is:

5t - 7g + 5x + 5gy = p

Reflection

Reflection is quite simple as it is a modification of translation. All that you need to do is to double the translation needed to take the shape to the line of symmetry. This formula will give you the T-total for the new shape. The part of the formula in red changes depending on what axis it is reflected in. If it is reflected in the y-axis then the part in red will be 5 + 7g see Rotation 1

5t - 7g + 2(5x + 5gy) = p

Rotation 1

In this section I'am going to investigate how the T-total and T-number change in r elation to each other, as they are rotated 90°, 180° and 270° from the original T-shape.

90°

To start with I am going to rotate the shape 90° when the point of rotation is t itself. Although they would overlap I have place the shapes separately so as to prevent confusion.

Position of the rotated shape in relation to the original.

As we can see because the shape is rotated about the T-number they both possess the original T-number. To find out the relationship between the T-total I have to add up all the terms within the new T-shape so as to obtain a formula that will work out the T-total.

t + ( t + 1 ) + ( t + 2 ) + ( t - ( g - 2 ) ) + ( t + ( g + 2 ) )

t + t + 1 + t + 2 + t - g + 2 + t + g + 2 = p

5t + 7 = p

We can see from this calculation that the new formula to find the T-total when the T-shape has been rotated 90° is very similar to the original. To see if this works for every T-shape which has been rotated 90° I will test a numerical example.

This is how the new rotation will appear.

Once again I will not place the two shapes as they would be but this is express in the diagram above.

If we now place the numbers into the formula we should find that the T-total is equal to 242.

5t + 7 = p

( 5 x 47 ) + 7 = 242

We now know that when a T-shape is rotated 90° around any given rotation point that the formula 5t + 7 = p will work to give the new T-total.

80°

The next step is to investigate the other two rotations of 180° and 270°. First however I will investigate 180°. Once again the small diagram expresses the position as the first rotation point will be t itself.

Once again the two share the same T-number, now I will add the terms together to get the new T-total.

t + ( t + g ) + ( t + 2g ) + ( t + ( 2g + 1 ) ) + ( t + ( 2g - 1 ) ) = p

t + t + g + t + 2g + t + 2g +1 + t + 2g - 1 = p

5t + 7g = p

What I can see from this is that it is the exact opposite to the original formula of the T-shape, 5t - 7g. To see if this works for every T-shape that has been rotated 180° I will test a numerical example.

This is how the new rotation will appear.

If we now place the numbers into the formula we should find that the T-total is equal to 395.

5t + 7g = p

( 5 x 65 ) + ( 7 x 10 ) = 395

We now know that when a T-shape is rotated 180° around any rotation point that the formula 5t + 7g = p will work to give the new T-total. It is important to note that the transformation of the original T-shape to 180° is in effect a reflection.

270°

The next step in investigating 270° I am going to shorten. Due to the formulae for 0° and 180° being opposite from each other I am going to make a prediction in saying that the formula for 270° will be opposite to that of 90°. If my prediction is correct then the formula 5t - 7 = p will work to find the T-total when the shape is rotated 270°. To clarify this I will first do an algebraic example to prove the formula followed by a numerical example to ensure that it works.

Once again the appearance of the main diagram will not be expressed accurately for practical reasons though the numbers and algebra will remain correct. What the rotation looks like will be shown in the smaller line diagram.

This is how the new rotation will appear.

If we now place the numbers into the formula we should find that the T-total is equal to 395.

t + ( t - 1 ) + ( t - 2 ) + ( t + ( -g - 2 ) ) + ( t + ( g - 2 ) ) = p

t + t - 1 + t - 2 + t - g - 2 + t + g - 2 = p

5t - 7 = p

My prediction was correct but now I need to ensure that the formula will work numerically as well as algebraically.

This is how the rotation will appear

If we now place the numbers into the formula we should find that the T-total is equal to 208.

5t - 7 = p

( 5 x 43 ) - ( 7 ) = 395

We now know that, as predicted, when a T-shape is rotated 270° around any rotation point that the formula 5t - 7 = p will work to give the new T-total.

Conclusion

The conclusion is that for rotation the following formulas work to find the T-total when the T-shape is rotated anywhere within the constraints of the grid.

Degrees rotated

Formulae

0°

5t - 7g = p *

90°

5t + 7 = p

80°

5t + 7g = p *

270°

5t - 7 = p

* Formulae of the same colour are opposite from one another.

Rotation 2

For the next stage of my investigation I am going to work out how to find the new T-number given the point of rotation and the original T-number. As I already have the formulae to find the T-total for each rotation these new formulae will be place in the previous formulae where t is present.

I am going to relate the point of rotation to the formula by including its grid position. For example in Diagram 1 the grid position of the point of rotation is (3,-1). It is this because I'am placing my point of origin as the original T-number. Establishing this, every square moved along the x-axis is equal to the x difference between the origin and the centre of rotation, in this case 3. It is the same with the y difference so that every square moved along the y-axis is, in this instance -1.

I am going to try several different rotations algebraically, display the results and then aim to find any correlations.

Diagram 1

* On this diagram (1) :

X = 3

Y = -1

What I am now going to do is say how I get from the original t to the new one in relation to the point of rotation.

t + 3 + g(y-3) = new t

t + 6 + g(2y) = new t

t + 4 + g(-2y) = new t

Diagram 2

* On this diagram (2) :

X = 0

Y = 2

What I am now going to do is say how I get from the original t to the new one in relation to the point of rotation.

t + 2 + g(y) = new t

t + 0 + g(2y) = new t

t + (-2) + g(y) = new t

Diagram 3

* On this diagram (3) :

X = 2

Y = 0

What I am now going to do is say how I get from the original t to the new one in relation to the point of rotation.

t + 2 + g( y - 2 ) = new t

t + 4 + g(y) = new t

t + 2 + g( y + 2 ) = new t

What I am now going to do is to see if there is any correlation between the three diagrams and each rotation.

90°

Diagram

Value of x

Value of y

T-number

3

-1

t + 2 + g( y - 3 )

2

0

2

t + 2 + g(y)

3

2

0

t + 2 + g( y - 2 )

* All of them begin with the original t so t will be part 1

* If you add x to y then it gives the middle term so ( y + x ) will be part 2

* To get the final term you take x from y and multiply by g so g( y - x) is part three

t + ( y + x ) + g( y - x ) = new t

This formula is to find the new t-total for any rotation of 90° given the original T-number and the point of rotation. To get the new T-total all you have to do is to apply the formulas done previously for rotation and T-total in Rotations 1. In this case:

5( t + ( y + x ) + g( y - x ) ) + 7 = new T-total

80°

Diagram

Value of x

Value of y

T-number

3

-1

t + 6 + g(2y)

2

0

2

t + 0 + g(2y)

3

2

0

t + 4 + g(y)

* All of them begin with the original t so t will be part 1

* If you multiply x by 2 you get the middle term so part 2 will be (2x)

* To get the final term you multiply y by 2 and then multiply that by g so that part three will be g( 2y )

t + ( 2x ) + g( 2y ) = new t

This formula is to find the new T-total for any rotation of 180° given the original T-number and the point of rotation. To get the new T-total all you have to do is to apply the formulas done previously for rotation and T-total in Rotations 1. In this case:

5( t + ( 2x ) + g( 2y ) ) + 7g = new T-total

270°

Diagram

Value of x

Value of y

T-number

3

-1

t + 4 + g(-2y)

2

0

2

t + (-2) + g(y)

3

2

0

t + 2 + g( y + 2 )

* All of them begin with the original t so t will be part 1

* If you subtract y from x you get the middle value so part 2 will be ( x - y )

* To get the final term you add x to y and multiply by g so g( y + x) is part three

t + ( x - y ) + g( y + x ) = new t

This formula is the to find the new t-total for any rotation of 90° given the original T-number and the point of rotation. To get the new T-total all you have to do is to apply the formulas done previously for rotation and T-total in Rotations 1. In this case:

5( t + ( x - y ) + g( y + x ) ) - 7 = new T-total

To conclude for Rotation

Rotation

Formula for new T-Total

90

5( t + (y+x) + g(y-x) ) + 7

80

5( t + 2x + g(2y) ) + 7g

270

5( t + (x-y) + g(y+x) ) - 7

Conclusion

In this investigation you can see how everything is linked together around the T-total and T-number. For example translation is involved in rotation and is modified for reflection.