Maths Evaluation for hidden fa
Evaluation
I have used a systematic approach when drawing the cubes and cuboids. I started with smaller cubes and increased their size. I increased one dimension at a time while holding the other two exactly the same. I started off by holding the width and height by one while increasing the depth by one and steadily going up in six, which is enough to give me an adequate set of results. In total, I increased each dimension (width, height the same.) I swapped separately while others where held still.
The dimensions that were held still were increased by going up In 3. However I increased the two inert dimensions by one after width, height and depth were increased separately.
Looking at my set of results let be w- width, H-height, D-depth, and n- the total numbers of cubes. After I have drawn the cubes for different dimensions I have worked out the number of cubes and hidden faces for each cube. For my results I have constructed formulas for the nth term for hidden faces of each six drawings but they only work for those dimensions.
My formula for the nth term for total faces in different dimensions is as the follow:
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Holding width and height by one and increasing the depth = 3n – 2
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Holding width and depth by one and increasing the height = 2n-1
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Holding height and depth by one and increasing the width = 3n – 2
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Holding width and height by two and increasing the depth = 4n – 2
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Holding width and depth by two and increasing the height = 4n – 4
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Holding height and depth by two and increasing the width = 4n + 2
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Holding the width and height by three and increasing the depth = 5n-18
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Holding width and depth by three and increasing the height = the formulas vary considerably as the size of cubes increases.
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Holding height and depth by three and increasing the width = 5n – 18
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Increasing all dimensions of a cube by and going up to six = the formulas vary considerably.
I found the formulas for particular dimensions of the cuboids. However some formulas are inconsistent as they vary with the size of the cubes. Each formula only works for a certain dimension numbers and it can’t be used to find the number of hidden faces in any cube. These formulas do not help me to find the total numbers of hidden faces in any cubes as they all keep hanging and varying. I cannot see a prospect of finding the general formula for the total number of hidden faces in any cube so I am going to stop drawing any more cubes. I am going to pot my results on the table (back page). In addition to the total number of cubes this time I will include the total number of faces, total number of faces seen on back and front, sides and top. The reason why I am going to include these values is because using only the total numbers of cubes is limiting my formula to a
Certain extent and is insufficient to find the general formula for the total number of hidden faces in any cubes so more values are required. After I plotted my table I will look for any patterns and values that will help me construct my ultimate formula. I have also fond out that the hidden faces have a pattern.
Conclusion
I have plotted a table with all my results to help me find the master formulas for the total number of hidden faces in any cubes. I have included dimensions (width, height, and depth), total number of cubes, total number of faces, and total number of faces on front, back, sides and top. I have also, constructed a formula for each value, they are as follow:
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Total numbers of cubes = width, height, depth
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Total numbers of faces = 6 (width, height, depth)
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Front and back = 2 (width, height)
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Sides = 2 ( height and depth)
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Top = (width and depth)
Now that I have found all the formulas I will try to construct the master formula, which would find the total numbers of hidden faces on any cube. Firstly, I will find a formula for all the visible faces on a cube, front and back, sides and top are only parts of a cube that are visible so the formula is 2 ( width, height) + 2 ( height and depth) + ( width and depth). Now that I have found the formula for the total numbers of visible faces I will need to find a formula for the total number of faces. The formula for the total number of cubes can be found by multiplying all three dimensions together. Then to find the total number of faces in any cube multiply the total numbers of cubes by six, therefore the formula for the total number of faces of any cube will, be 6 (width, base height). If I join both formulas I will get the master formula for the total number of hidden faces of any cube. As a result the formula is 6 (width, base, height) – (2 width, height, +2 height, and depth).
Now to see if my formula works I am going to put it a test. I am going to put it as a test. I am going to select two random dimensions and use my formula to find the total numbers of hidden faces. Then I will check if my formula has worked. The first dimensions that I would use is width = 5, height = 1 and depth = 1. 6 (5, 1, 1) = 30 and (2, 5,1+2,1,1+5,5) = 17. Therefore 30-17=13 which is correct. Width=5, height = 5, and depth = 5. 6 (5,5,5)= 750 and (2,5,5,+2,5,5,+5,5) = 125 so 750- 125 = 625, which again show I am on the right way. Now my formula is consistent as it works on any dimension of a cube and finds the exact number of hidden faces. This could not be done by a formula containing only the total numbers of cubes as it limits the formula so it works only on a particular dimension. However some formulas that use only the total numbers of cubes do not work on particular dimensions at all keep varying as the size of cube increases because the formula has an insufficient amount of values. The main reason why I found the formula was because I split up my cube into different values such as sides, front and back and top, which were crucial in helping me to find the total numbers of seen faces and the rest seemed to have followed as it was rather easy to find the total number of cubes and faces. Now that I have found the general formula I have worked out further hidden faces in different dimensions and put my results on the table.
