Chapter 11.6, Problem 11P

(a)

To determine

To find:the value and fill the blanks in table form Koch curve.

Answer to Problem 11P

Number of new squares	$1$	$4$	$20$	$100$	………….	${4.5}^{n-2}$
Area of each new square	$1$	$\frac{1}{9}$	$\frac{1}{81}$	$\frac{1}{9^3}$	…………..	$\frac{1}{9^{n-1}}$
Total new area	$1$	$\frac{4}{9}$	$\frac{20}{81}$	$\frac{100}{9^3}$	…………..	$\frac{{4.5}^{n-2}}{9^{n-1}}$

Explanation of Solution

Given:

  

Concept used:

The geometric progression:

  $a,ar,a{r}^2\mathrm{.....}a{r}^{n-1}.$

Last term or nth term of the geometric progression will be:

  $t_n=a{r}^{n-1}$ .

  $s_n=\frac{a\left(1-r^n\right)}{1-r};s_{\infty }=\frac{a_1}{1-r}$ .

Calculation:

the table so form is

	Fig. $1$	Fig. $2$	Fig. $3$	Fig. $4$	…………..	Fig. $n$
Numberofnew squares	$1$	$4$	$20$	$100$	………….	${4.5}^{n-2}$
Areaofeach new square	$1$	$\frac{1}{9}$	$\frac{1}{81}$	$\frac{1}{9^3}$	…………..	$\frac{1}{9^{n-1}}$
Total new area	$1$	$\frac{4}{9}$	$\frac{20}{81}$	$\frac{100}{9^3}$	…………..	$\frac{{4.5}^{n-2}}{9^{n-1}}$

(b)

To determine

To find: the total area bounded by the Koch curve.

Answer to Problem 11P

Area is $2$ .

Explanation of Solution

Given:

  

Concept used:

The geometric progression:

  $a,ar,a{r}^2\mathrm{.....}a{r}^{n-1}.$

Last term or nth term of the geometric progression will be:

  $t_n=a{r}^{n-1}$ .

  $s_n=\frac{a\left(1-r^n\right)}{1-r};s_{\infty }=\frac{a_1}{1-r}$ .

Calculation:

The table for the Koch curve is:

	Fig. $1$	Fig. $2$	Fig. $3$	Fig. $4$	…………..	Fig. $n$
Number of new squares	$1$	$4$	$20$	$100$	………….	${4.5}^{n-2}$
Area of each new square	$1$	$\frac{1}{9}$	$\frac{1}{81}$	$\frac{1}{9^3}$	…………..	$\frac{1}{9^{n-1}}$
Total new area	$1$	$\frac{4}{9}$	$\frac{20}{81}$	$\frac{100}{9^3}$	…………..	$\frac{{4.5}^{n-2}}{9^{n-1}}$

The area of the square can be calculated as:

Area= $1+\frac{4}{9}{\displaystyle {\sum }_{k=0}^{\infty }{\left(\frac{5}{9}\right)}^k}=1+\frac{4}{9}\left(\frac{1}{1-\frac{5}{9}}\right)=2.$

Hence, area is $2$ .