Math Coursework

Math Coursework

Math Coursework (Anyone for T) T Shapes (T Values)

To investigate the behaviours of the T values we need to first have a large sample of calculated T values to analyze and create a general method in which the T values can be calculated, we will get the T values for numbers 1 to 20.

T1

222 = 484

1×3=3

=484–3=481

T2

232 = 529

2×4=8

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Math Coursework

=529–8=521

T3

242 = 576

3×5=15

=576–15=561

T4

252 = 625

4×6=24

= 625 – 24 = 601

T5

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Math Coursework

262 = 676

5×7=35

=676–35=641

T6

272 = 729

6×8=48

=729–48=681

T7

282 = 784

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Math Coursework

7×9=63

=784–63=721

T8

292 = 841

8×10=80

=841-80=761

T11

322 = 1024

11×13=143

= 102 –143 = 881

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Math Coursework

T12

332 = 1089

12×14=168

1089 – 168 = 921

T13

342 = 1156

13×15=195

1156 – 195 = 961

T14

352 = 1225

14×16=224

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Math Coursework

=1225–224= 1001

T15

362 = 1296

15×17=255

1296 –255 = 1041

T16

372 = 1369

16×18=288

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Math Coursework

1369 –288 = 1081

T17

382 = 1444

17×19=323

1444 –323 = 1121

T18

392 = 1521

18×20=360

1521 –360 = 1161

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Math Coursework

From the above results an observable pattern is that the results for the T value for each number is the previous value of T plus 40. Therefore we could formulate a formula as follows:

T n = Tn-1 + 40

The T values can be derived with the following formula where we use T1 to get all the other

values, the value for T 1 is 481, therefore we can get a general formula as follows:

T n = 481 + (40 × (n-1))

We can test the validity of this formula by using the already calculated values of T, for example we will use T17 to verify this as follows:

T17 = 481 (40 × (17- 1))

Our results will be as follows

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Math Coursework

T17 = 481 + (40 × 16)

T17 = 481 + 640

= 1121

This is the same value we got when we used the other method, we will test the formula again using T16 as follows:

T16 = 481 + (40 × (16 – 1))

T16 = 481 + (40 × 15)

= 481 + 600

=1081

This is the same value we got from our earlier calculations and therefore our formula that

T n = 481 + (40 × (n-1)) qualifies as the general formula to get the T values for any value in our case.

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Math Coursework

The T values can be graphically represented in the graph above.

If we were to shift the T side ways this will result into an increase in the value of T by 40, if in the other case we shift the T downwards there will be an increase by four hundred, example a shift from T13 to T23, T 13 is 961 while T23 is 1361, this is an increase by 400, this can be linked to our formula in that the change in T is 23 – 13 which gives us 10, one step movement of the T is 40 and therefore 40 x 10 gives us 400.

We could also use algebra to formulate a general formula to get the value of T; these will be done through the following procedure:

Due to the fact that T13 is calculated through getting the square of 34 and then subtracting the product of 13 and 15 we can use these figures to formulate an algebraic formula that give us the general formula;

We can first assume that n represent 13, therefore the t value will be calculated by the following formula:

. ((n + 1) +20)2 – (n × (n + 2))

= (n+21)2 – (n2 +2n)

= (n2 +42n + 441) – (n2 +2n)

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Math Coursework

= n2-n2+42n -2n +441

= 40n + 441

Therefore our T value can be simply calculated by the simple formula:

T n = 40n+441

We can prove this formula by calculating the value of T1

T1= 40(1) + 441

= 481

ROTATION:

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Math Coursework

We can investigate what happens to T values ones we rotate the T to 90,180 and 270 degrees,

90 degrees rotation (clockwise rotation)

When we rotate the T at 90 degrees the new T value assumes the following formula,

T n = (n+10)2 – ((n+2) x (n+22))

= (n2+20n + 100) – (n2 +24n +44)

Therefore our general formula for T when there is a 90 degree rotation is

T n = – 4n +54 or T n =54 – 4n

If we compare this value with the original T which assumed the formula T n = 40n+441, the difference is 40n+441-(54 – 4n) which gives us

36n +387

180 degrees rotation:

The new T will take the following formula:

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Math Coursework

T n = (n+1)2 – ((n+22) x (n+20))

= (n2 + 2n + 1) – (n2 +42n+440) Therefore our formula for T is

T n = – 439 – 40n

If we compare this value with the original T formula T n = 40n+441 the difference will be

40n +441- (-439 – 40n) = 80n + 880

270 degree rotation

In this case the T formula will below as follows:

T n = (n+12)2 – (n (n+20))

= (n2 +24n + 144) – (n2 +20n)

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Math Coursework

T n = 4n +144

If we compare this value with the original value of T which is T n = 40n+441, then the difference will be 40n+441-(4n+144) = 36n +297

We can therefore conclude that the rotation of T does not have any significant formula to analyse the rotation behaviour of T values because the rotation formulas do not have any significant pattern.

ENLARGMENT:

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Math Coursework

We will investigate what happens when we enlarge the T

When we enlarge the figure it takes the above form, therefore our new T value will be

T n = (n+31)2 – ((n-1) x (n+3))

= (n2 + 62n + 961) – (n2 +2n -3) Therefore the value of our T will be

T n =60n +964

When we compare these value of T with the original T value T n = 40n+441 then our difference will be 40n+441 – (60n + 964) = -20n – 523

The T value however as a result of enlargement does not have any significant mathematical formula, we expected that the T value would have doubled or even be a square of the original value, however this is not the case but the formulas calculated will give the t value for enlargement for any T value.

References:

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Math Coursework

Math Coursework (2007) Retrieved on 1st march, Available at http://www.ocr.org.uk/Data/publi cations/specifications_syllabuses_and_tutors_handbooks

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