Calculus Projects with Maple

Project 20. Koch's Fractal

1. Background
2. Maple Code
3. Project Report
4. Extension
5. Requirements

For this project you will need familiarity with the commands:

In addition, you will need to use the style and numpoints options of plot and the two argument form of arctan.

Background

r3 := sqrt(3);
snow_0 := [[0,r3/6], [1/2,r3*2/3], [1,r3/6], [0,r3/6]]:
plot(snow_0, scaling=constrained);

Remove the middle third of each side and insert two sides of a smaller equilateral triangle as shown below.

snow_1 := [[0,r3/6], [1/6,r3/3], [0,r3/2], [1/3, r3/2], [1/2, 2*r3/3],
   [2/3, r3/2], [1, r3/2], [5/6, r3/3], [1,r3/6], [2/3,r3/6], [1/2,0],
   [1/3,r3/6], [0,r3/6]]:
plot(snow_1, scaling=constrained);

There are two observations to make:

1. Since the length of each side has increased by one third, the total perimeter has increased by one third.
2. The island is still contained in a rectangle of size 1 by 2*3^(1/2)/3.

I. Calculate:

1. The perimeters of the original and first snowflakes.
2. The areas of the original and first islands.
3. The distance from a new vertex in the first snowflake to the edge of the original snowflake.

The second iteration is shown below. Try it---if you're up to the typing (or use copy and paste ).

snow_2 :=
 [ [0,    r3/6],    [1/18,r3*2/9],   [0,    r3*5/18],  [1/9,  r3*5/18],
   [1/6,  r3/3],    [1/9, r3*7/18],  [0,    r3*7/18],  [1/18, r3*4/9],
   
   [0,    r3/2],    [1/9, r3/2],     [1/6,  r3*5/9],   [2/9,  r3/2],
   [1/3,  r3/2],    [7/18,r3*5/9],   [1/3,  r3*11/18], [4/9,  r3*11/18],
   
   [1/2,  r3*2/3],  [5/9, r3*11/18], [2/3,  r3*11/18], [11/18,r3*5/9],
   [2/3,  r3/2],    [7/9, r3/2],     [5/6,  r3*5/9],   [8/9,  r3/2],

   [1, r3/2],

   [17/18,r3*4/9],  [1,   r3*7/18],  [8/9,  r3*7/18],  [5/6,  r3/3],
   [8/9,  r3*5/18], [1,   r3*5/18],  [17/18,r3*2/9],   [1,    r3/6],
   
   [8/9,  r3/6],    [5/6, r3/9],     [7/9,  r3/6],     [2/3,  r3/6],
   [11/18,r3/9],    [2/3, r3/18],    [5/9,  r3/18],    [1/2,  0],
   
   [4/9,  r3/18],   [1/3, r3/18],    [7/18, r3/9],     [1/3,  r3/6],
   [2/9,  r3/6],    [1/6, r3/9],     [1/9,  r3/6],     [0,    r3/6] ]:
   
plot(snow_2, scaling=constrained);

For those who aren't up to the typing, a printout of a Maple session with the routines for generating and animating Koch snowflakes follows.

Maple Code

Note: You must use Maple V Release 2 or a newer version to use the routines below. If you have an earlier version then jump to the bottom of this section and click on the QuickTime movie.

####################################
##  Animating the Koch Snowflake  ##
##  (requires Maple V Release 2)  ##
##  (or newer version          )  ##
####################################

# setup plotting
plots[setoptions](axes=none, scaling=constrained);

##############################################
## A proc to "connect the dots" to plot a path
## (mainly for Maple V Release 2)
##
graph := proc() local pts;
    pts := [args];
    plot(pts, style= LINE, numpoints=nargs)
    end:
  
##################################################
## Rotate point A about the origin through angle p
##
rotate := proc(a, p);
    [p[1]*cos(a)-p[2]*sin(a), p[1]*sin(a)+p[2]*cos(a)];
    end:
  
###########################################################
## Insert three points of a Koch triangle between pts A & B
##
Koch := proc(A, B)   local pA, pB, pC, dx, dy, len, alpha, delta;

    pA := [(2/3*A[1]+1/3*B[1]), (2/3*A[2]+1/3*B[2])];

    dx    := B[1]-A[1];
    dy    := B[2]-A[2];
    len   := sqrt(dx^2+dy^2)/3;
    alpha := arctan(dy, dx);   # Note the comma
    delta := rotate(alpha, [len/2, sqrt(3)*len/2]);

    pB := [(pA[1]+delta[1]), (pA[2]+delta[2])];
    pC := [(1/3*A[1]+2/3*B[1]) ,(1/3*A[2]+2/3*B[2])];

    pA, pB, pC;
    end:

#######################################################
## Take a path and iterate with inserted Koch triangles
##
path := proc()  local pa, pth, i;
    pa := args;
    pth := pa[1];
    for i from 2 to nops([pa])
      do
        pth := (pth, Koch(pa[i-1], pa[i]), pa[i]);
      od;
    pth:
    end:

## Start with an equilateral triangle #
s := sqrt(3)/6:
  p := [0, s], [1/2, 4*s], [1, s], [0, s]:
  snow.0 := graph(p):

## Now begin the iterations - go up to 5
N := 5:
  for i from 1 to N
    do
      p := path(p);
      snow.i := graph(p);
	od:

## and animate
plots[display]([snow.(0..N)], insequence=true);

#################################################

For Maple 5.0 users Look at the QuickTime movie (on a Macintosh).

II. Calculate:

1. The perimeter of the second snowflake.
2. The area of the second island.
3. The distance from a new vertex in the second snowflake to the edge of the first snowflake.

For each successive iteration, repeat the procedure of inserting smaller and smaller triangles as seen in the figure below.

	The First Three Iterations

How many line segments are in the n -th iteration's curve?

The Koch Snowflake is the limit curve of this procedure. The snowflake has the remarkable property that at no point does the curve possess a tangent line. To see the curve develop, run the animation using the Maple procedures given in the listing above in this project.

III. Determine:

1. At which iteration can you no longer see a difference in the image displayed on the screen?
2. On a laser printout?

Project Report

1. Develop a formula L(n) that gives the perimeter of S(n), the n -th snowflake. Find the limit of the sequence L(n), that is, find:
   limit(L(n), n=infinity).

2. Develop a formula A(n) that gives the area enclosed by S(n). Find:
   limit(A(n), n=infinity).

3. Develop a formula D(n) that gives the distance from a new vertex of S(n) to the edge of S(n-1). Find:
   limit(D(n), n=infinity).

4. Discuss the apparent paradox.

Extension

Requirements

1. Answers to Questions 1 through 4.
2. Full responses to Problems 1 and 2 with supporting arguments.