Chapter 11.6, Problem 13P

To determine

To find: The total area bounded by the snowflake curve.

Answer to Problem 13P

The total area bounded the snowflake is $\frac{2\sqrt{3}}{5}\text{sq}\text{units}$ .

Explanation of Solution

Given information:

The snowflake curve.

  

Let, the area of snowflake be $A_k$ .

Let, the number of sides of snowflake be $N_k$ .

Let, the length of the sides of snowflake be $L_k$ .

The area of equilateral triangle can be calculated by the formula,

  $A=\frac{\sqrt{3}}{4}{s}^2$ .

The variables above are initialized as follows:

  $A_k=a,N_k=3,L_k=x$ .

Since the snowflake is an equilateral triangle. It can be easily seen that the length of each side is reduced by a factor three with each iteration.

Therefore,

  $L_{k+1}=\frac{l_k}{3},L_o=x,L_K=\frac{x}{3^k}$ .

The number of sides is increased by factor $4$ . Therefore,

  $N_{k+1}=4n_k$ .

The area of snowflake is,

  $A_{k+1}=A_k+N_k\alpha \left(l_{k+1}\right)$ .

For $A=0$ ,

  $\begin{array}{c}{A}_1=a_0+N_0\left(L_1\right)\\ =a\left(x\right)+3a\left(\frac{x}{3}\right)\end{array}$

For $A=1$ ,

  $\begin{array}{c}{A}_2=a_1+N_1\left(L_2\right)\\ =a\left(x\right)+3a\left(\frac{x}{3}\right)+4\cdot 3a\left(\frac{x}{3^2}\right)\end{array}$

For $A=2$ ,

  $\begin{array}{c}{A}_3=a_1+N_{2}a\left(L_3\right)\\ =a\left(x\right)+3a\left(\frac{x}{3}\right)+4\cdot 3a\left(\frac{x}{3^2}\right)+4^2\cdot 3a\left(\frac{x}{3^3}\right)\end{array}$

Finally,

  $\begin{array}{c}{A}_k=a\left(x\right)+\underset{m=1}{\overset{k}{{\displaystyle \Sigma }}}{4}^{m-1}\cdot 3a\left(\frac{x}{3^m}\right)\\ =a\left(x\right)+\frac{3}{4}\underset{m=1}{\overset{k}{{\displaystyle \Sigma }}}{4}^m\cdot a\left(\frac{x}{3^m}\right)\\ =3^{-2m}a\left(x\right)\\ =9^{-m}a\left(x\right)\end{array}$

The area of snowflake is,

  $A=a\left(x\right)\left(1+\frac{3}{4}{\displaystyle \sum _{m=1}^k{\left(\frac{4}{9}\right)}^m}\right)$ .

The sum of $k$ terms with common ration is,

  $\begin{array}{c}{{\displaystyle \sum _{m=1}^k\left(\frac{9}{4}\right)}}^m=\frac{4}{9}÷1-\frac{4}{9}{\left(1-\frac{4}{9}\right)}^k\\ =\frac{4}{5}\left(1-{\frac{4}{9}}^m\right)\end{array}$

Therefore, the area of snowflake is $A=a\left(x\right)\left(\frac{8}{5}-\frac{3}{5}{\left(\frac{4}{9}\right)}^k\right)$ .

The area of Koch snowflake is given by taking the limit. Thus,

  $\begin{array}{c}\underset{k\to \infty }{{\displaystyle \lim }}{A}_k=\frac{8}{5}a\left(x\right)\\ \underset{k\to \infty }{{\displaystyle \lim }}{A}_k=\frac{8}{5}\frac{\sqrt{3}}{4}{x}^2\\ =\frac{2\sqrt{3}}{5}{x}^2\end{array}$

Therefore, the total area bounded the snowflake is $\frac{2\sqrt{3}}{5}\text{sq}\text{units}$ .