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Teachers Notes This is a brief but very interesting look, at the Von Koch Snowflake Curve. After introducing the curve and discussing its generation, the students are simply asked to derive the perimeter formula for nth iteration. We then move on to discuss the curve’s finite area and reveal, (by reference to the formula) its infinite perimeter. Students are encouraged to generate a spreadsheet from the formula for the first 50 terms in the sequence to convince themselves of the infinity of the perimeter. (Spreadsheet is at slide 16). There is a printable worksheet if needed at slide 18 (some students may wish to make jottings/notes on it). If you want to extend still further for the very able, then you might wish to see if they can work out the area An for the nth iteration.
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The Von Koch Snowflake
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The Von Koch Snowflake
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The Von Koch Snowflake
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The Von Koch Snowflake
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The Von Koch Snowflake
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The Von Koch Snowflake P0 = L 1 3 L 1 3 L 1 3 L
The curve is generated from an equilateral triangle by trisecting the sides and constructing this smaller equilateral triangle on each of the sides. This is then repeated ad infinitum. 1 3 L
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The Von Koch Snowflake P0 = L P1 = 4 3 L 1 3 L 1 3 L 1 3 L
Thinking about the increased length of this side, what will the first new perimeter, P1 be? 1 3 L
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The Von Koch Snowflake P0 = L P1 = 4 3 L P2 =( )2 4 3 L 1 3 L 1 3 L 1
Derive a general formula for the perimeter of the nth curve in this sequence, Pn. 1 3 L
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The Von Koch Snowflake P0 = L P1 = 4 3 L P2 =( )2 4 3 L 1 3 1 3
Derive a general formula for the perimeter of the nth curve in this sequence, Pn. Pn =( )n 4 3 L 1 3
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P1 = 4 3 L P0 = L A0 A1 A2 A3 P2 =( )2 4 3 L P3 =( )3 4 3 L
The area An of the nth curve is finite. This can be seen by constructing the circumscribed circle about the original triangle as shown. A2 A3 P2 =( )2 4 3 L P3 =( )3 4 3 L
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P1 = 4 3 L P0 = L A0 A1 Pn =( )n 4 3 L A2 A3 P2 =( )2 4 3 L P3 =( )3 4
It is a surprising fact therefore that the perimeter of the curve is infinite. Pn =( )n 4 3 L A2 A3 P2 =( )2 4 3 L P3 =( )3 4 3 L
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P1 = 4 3 L P0 = L A0 A1 Pn =( )n 4 3 L A2 A3 P2 =( )2 4 3 L P3 =( )3 4
Whatever fixed value you care to make the perimeter of any curve in the sequence it can always be exceeded by choosing a large enough value for n. Pn =( )n 4 3 L A2 A3 P2 =( )2 4 3 L P3 =( )3 4 3 L
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P1 = 4 3 L P0 = L A0 A1 Pn =( )n 4 3 L A2 A3 P2 =( )2 4 3 L P3 =( )3 4
Use a spreadsheet to compute the first 50 values for the perimeter. Set P0 = 1. Pn =( )n 4 3 L A2 A3 P2 =( )2 4 3 L P3 =( )3 4 3 L
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P0 1 P1 1.333 P26 P2 1.778 P27 P3 2.370 P28 P4 3.160 P29 P5 4.214 P30 P6 5.619 P31 P7 7.492 P32 P8 9.989 P33 P9 13.318 P34 P10 17.758 P35 P11 23.677 P36 P12 31.569 P37 P13 42.092 P38 P14 56.123 P39 P15 74.831 P40 P16 99.775 P41 P17 P42 P18 P43 P19 P44 P20 P45 P21 P46 P22 P47 P23 P48 P24 P49 P25 P50
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The Von Koch Snowflake The perimeter of the Von Koch Snowflake Curve is infinite. Just as the coast line of the UK is infinite. The smaller the ruler that you use to measure the coast line, the longer it becomes. Coast line miles
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