Chapter 8, Problem 3P

(a)

To determine

To Find: The formula for $s_n,l_n$ and $p_n$ .

Answer to Problem 3P

The required formula is $s_n=3\cdot 4^n$ , $L_n=3\cdot 4^n{\left(\frac{1}{3}\right)}^n$ and $p_n=3\cdot {\left(\frac{4}{3}\right)}^n$ .

Explanation of Solution

Given:

The given diagram is shown in Figure 1

  

Figure 1

Calculation:

Consider the table shown in Table 1

Table 1

$s_0=3$	$L_0=1$
$s_1=3\cdot 4$	$L_1=\frac{1}{3}$
$s_2=3\cdot 4^2$	$L_2={\left(\frac{1}{3}\right)}^2$
$s_3=3\cdot 4^3$	$L_3={\left(\frac{1}{3}\right)}^3$

Consider the stage each side is replaced by the shorter sides each of the length of $\frac{1}{3}$ and the side length at the preceding stage. Then, write $s_0$ and $L_0$ for the number of sides and the length of the sides of the initial triangle and then generate the table for $s_n=3\cdot 4^n$ and $L_n=3\cdot 4^n{\left(\frac{1}{3}\right)}^n$ this, shows that the length of the perimeter at the nth stage of the construction is,

  $\begin{array}{c}{p}_n=s_n{L}_n\\ =3\cdot {\left(\frac{4}{3}\right)}^n\end{array}$

Thus, the required formula is $s_n=3\cdot 4^n$ , $L_n=3\cdot 4^n{\left(\frac{1}{3}\right)}^n$ and $p_n=3\cdot {\left(\frac{4}{3}\right)}^n$ .

(b)

To determine

To Prove: The value $p_n\to \infty$ and $n\to \infty$ .

Explanation of Solution

Calculation:

Consider the formula for $p_n$ is,

  $p_n=4{\left(\frac{4}{3}\right)}^{n-1}$

Since, $\frac{4}{3}>1$ then, $p_n\to \infty$ as $n\to \infty$ .

Hence, proved.

(c)

To determine

To Find: The sum of the infinite series to find the area enclosed by the snow flake curve.

Answer to Problem 3P

The area enclosed by the snowflake is $\frac{2\sqrt{3}}{5}$ .

Explanation of Solution

Calculation:

The area of each of the small triangles added at the given stage is one fourth of the area of the triangle added at the preceding stage. Consider the area of the original triangle and then, the area $a_n$ of each of the small triangles added at the stage n is,

  $a_n=\frac{a}{9^n}$

Since, the small triangle is added to the sides at every stage so it follow that the area is added to the figure of the next stage as,

  $\begin{array}{c}{A}_n=S_{n-1}{a}_n\\ =3\cdot 4^{n-1}\cdot \frac{a}{9^n}\\ =a\frac{4^{n-1}}{3^{2n-1}}\end{array}$

Then, the total area is of the form,

  $\begin{array}{c}A=a+A_1+A_2+\mathrm{....}\\ A=a+a\frac{1}{3}+a\frac{4}{3^3}+a\frac{4^2}{3^5}+\mathrm{..}\\ A=a+\frac{\frac{a}{3}}{1-\frac{4}{9}}\\ A=\frac{8a}{5}\end{array}$

Then, the area of the original equilateral triangle with sides is,

  $\begin{array}{c}a=\frac{1}{2}\cdot 1\sin \left(\frac{\pi }{3}\right)\\ =\frac{\sqrt{3}}{4}\end{array}$

Thus, the area of the enclosed by the snowflake curve is,

  $\frac{8}{5}\left(\frac{\sqrt{3}}{4}\right)=\frac{2\sqrt{3}}{5}$