# GCSE Maths Bad Tomato Investigation

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Introduction

GCSE Maths Bad Tomato Investigation A problem has been reported involving Bad tomatoes. At first there is a tray of regular tomatoes, then, all of a sudden, one of them goes bad. An hour later, all of the ones that were touching the bad tomato go bad. An hour after that all the tomatoes touching those go bad. This continues until the whole tray goes bad. To start this investigation I am going to use a simple 4x4 box. Let's say, after one hour, a tomato goes bad. This is marked 1. After the second hour all those touching it go bad, those are marked 2 and so on. 2 3 4 5 1 2 3 4 2 3 4 5 3 4 5 6 With a 4x4 tray, the time taken from the tomatoes being put into the tray to the last tomato going rotten is 6 hours. Hour No. of bad toms after that hour No. of tomatoes that turn bad that hour 0 0 0 1 1 1 2 4 3 3 8 4 4 12 4 5 15 3 6 16 1 A pattern is beginning to form in the "No. of tomatoes that turn bad that hour" column. If we continue this investigation to larger sized trays then we will see that the pattern becomes even more prominent. 2 3 4 5 6 7 8 9 10 11 1 2 3 4 5 6 7 8 9 10 2 3 4 5 6 7 8 9 10 11 3 4 5 6 7 8 9 10 11 12 4 5 6 7 8 9 10 11 12 13 5 6 7 8 9 10 11 12 13 14 6 7 8 9 10 11 12 13 14 15 ...read more.

Middle

from X is 5 away so D2=5 +1 for the fist tomato to go bad T=13 A formula using the co-ordinates of X would make the calculation allot easier. If we now try a 10x10 grid starting at (7,5) 12 11 10 9 8 7 6 10 8 9 11 10 9 8 7 6 5 6 7 8 10 9 8 7 6 5 4 5 6 7 9 8 7 6 5 4 3 4 5 6 8 7 6 5 4 3 2 3 4 5 7 6 5 4 3 2 1 2 3 4 8 7 6 5 4 3 2 3 4 5 9 8 7 6 5 4 3 4 5 6 10 9 8 7 6 5 4 5 6 7 11 10 9 8 7 6 5 6 7 8 D1+D2+1=T still works for this starting position. The affects of the co-ordinates also seems to follow a logical pattern. T can be worked out by using the following program. We will call the first co-ordinate C1 and the second C2. T=0 If C1 < 1/2L then T=T+(L-C1) If C1 > 1/2L then T=T+C1 If C2 < 1/2L then T=T+(L-C2) If C2 > 1/2L then T=T+C2 T=T+1 We can work this through on a 100x100 grid starting point (97,43) 100/2 = 50 97 is more than 50 so T = 97 43 is less than 50 so T = 97 + (100-43) + 1 T = 155 I am not going to draw out a 100x100 grid but this does seem a very sensible answer. If we know that the tomato that goes bad will always be in the corner then we could investigate a formula for an infinitely big tray. ...read more.

Conclusion

The last thing we will look at is infinitely big 3D shapes. This growth should be very similar to that of a 2D shape; only an extra part will need to be added to take account of the vertical expansion. Hour (H) No. of tomatoes going bad at hour H (B) 2 6 3 18 4 38 5 66 6 102 As I thought, it is very similar to the 2D pattern. The formula for this would be 4(H-1)2+2=B (H is squared to take into account the extra dimension). From this we can work out how to find the number of tomatoes that have gone bad after a given number of hours in an infinite cube, and from that adapt the 2D formulas to take into account starting position. Hour (H) Total No. of tomatoes gone bad by hour H 2 7 3 25 4 63 5 129 6 231 The only way I can find to calculate the number of tomatoes that have gone bad after a certain time is to work it out using the 4(H-1)2+2=B formula. I am sure there is a formula to work out the spread in a 3D shape, but I am also sure that it is a very complex one, either that or I am missing something simple. H3 is very close but all the results are slightly off H H^3 Actual Difference 2 8 7 -1 3 27 25 -2 4 64 63 -1 5 125 129 4 6 216 231 15 7 343 377 34 8 512 575 63 9 729 833 104 10 1000 1159 159 11 1331 1561 230 12 1728 2047 319 13 2197 2625 428 14 2744 3303 559 15 3375 4089 714 Well, it looks like there is a pattern in the Difference column, but what it is? ...read more.

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