Chapter 7.3, Problem 2MRPS

Solution

a.

To calculate : The rule for the area of $n^{th}$ ring if the target has rings that are each 1 foot wide as shown in the figure below.

  

The rule for the area of $n^{th}$ ring is $\left(2n-1\right)\pi$ .

Given information :

The target has rings that are each 1 foot wide as shown in the figure below.

  

Formula used :

The $n^{th}$ term of a arithmetic series with first term as $a_1$ and common difference $d$ is given by $a_n=a_1+\left(n-1\right)d$ .

Calculation :

Consider the target ring as shown below.

  

Calculate the area of the first ring $a_1$ , the bulls’ eyes circle using the area of circle of the radius $1$ ft.

  $\begin{array}{c}A=\pi r^2\\ =\pi {\left(1\right)}^2\\ =\pi \end{array}$

  $a_1=\pi$

Calculate the area of the second ring $a_2$ that lies between the second and first circle.

  $\begin{array}{c}{a}_2=\pi {\left(2\right)}^2-\pi {\left(1\right)}^2\\ =4\pi -\pi \\ =3\pi \end{array}$

Similarly, calculate the area of the third ring $a_3$ as follows.

  $\begin{array}{c}{a}_3=\pi {\left(3\right)}^2-\pi {\left(2\right)}^2\\ =9\pi -4\pi \\ =5\pi \end{array}$

The series makes an arithmetic series with $a_1=\pi$ and $d=2\pi$ .

Write the arithmetic series as follows.

  $\begin{array}{c}{a}_n=\pi +\left(n-1\right)2\pi \\ =\pi +2n\pi -2\pi \\ =2n\pi -\pi \\ =\left(2n-1\right)\pi \end{array}$

Thus, the rule for the area of $n^{th}$ ring is $\left(2n-1\right)\pi$ .

b.

To calculate : The summation notation for the series giving total area of target with $n$ rings if the target has rings that are each 1 foot wide as shown in the figure below.

  

The rule for the area of $n^{th}$ ring is $\left(2n-1\right)\pi$ .

Given information :

The target has rings that are each 1 foot wide as shown in the figure below.

  

Formula used :

The sum of first $n$ positive integers is calculated as $\frac{n\left(n+1\right)}{2}$ .

Calculation :

Consider the rule for the area of $n^{th}$ ring as $\left(2n-1\right)\pi$ .

The summation of $n$ terms can be written as follows.

  $\begin{array}{c}{\displaystyle \sum _{i=1}^n{a}_i}={\displaystyle \sum _{i=1}^n\left(2n-1\right)\pi }\\ ={\displaystyle \sum _{i=1}^{n}2n\pi }-{\displaystyle \sum _{i=1}^n\pi }\end{array}$

Use the formula for the sum of first $n$ positive integers.

  $\begin{array}{c}{\displaystyle \sum _{i=1}^n{a}_i}=2\pi \left(\frac{n\left(n+1\right)}{2}\right)-n\pi \\ =\pi \left(n^2+n\right)-n\pi \\ =n^2\pi +n\pi -n\pi \\ =n^2\pi \end{array}$

Thus, summation notation for the series giving total area of target with $n$ rings is $n^2\pi$ .

c.

To calculate : The value of the expression from part (b) when $n=1,2,4,8$ and the effect of doubling the number of rings have on total area of target with $n$ rings if the target has rings that are each 1 foot wide as shown in the figure below.

  

The value of the expression from part (b) when $n=1,2,4,8$ is $\pi ,4\pi ,16\pi ,64\pi$ and doubling the number of rings make the area of the target $4$ times bigger.

Given information :

The target has rings that are each 1 foot wide as shown in the figure below.

  

Formula used :

Substitute the desired values in the given expression.

Calculation :

From part (b) consider, the sum ${\displaystyle \sum _{i=1}^n\left(2n-1\right)\pi }$ .

For $n=1$ ,

  $\begin{array}{c}{\displaystyle \sum _{i=1}^1\left(2n-1\right)\pi }=\left(2\left(1\right)-1\right)\pi \\ =\left(2-1\right)\pi \\ =\pi \end{array}$

For $n=2$ ,

  $\begin{array}{c}{\displaystyle \sum _{i=1}^2\left(2n-1\right)\pi }=\left(2\left(1\right)-1\right)\pi +\left(2\left(2\right)-1\right)\pi \\ =\left(2-1\right)\pi +\left(4-1\right)\pi \\ =\pi +3\pi \\ =4\pi \end{array}$

For $n=4$ ,

  $\begin{array}{c}{\displaystyle \sum _{i=1}^4\left(2n-1\right)\pi }=\left(2\left(1\right)-1\right)\pi +\left(2\left(2\right)-1\right)\pi +\left(2\left(3\right)-1\right)\pi +\left(2\left(4\right)-1\right)\pi \\ =\left(2-1\right)\pi +\left(4-1\right)\pi +\left(6-1\right)\pi +\left(8-1\right)\pi \\ =\pi +3\pi +5\pi +7\pi \\ =16\pi \end{array}$

For $n=8$ ,

  $\begin{array}{c}{\displaystyle \sum _{i=1}^8\left(2n-1\right)\pi }=\left(2\left(1\right)-1\right)\pi +\left(2\left(2\right)-1\right)\pi +\left(2\left(3\right)-1\right)\pi +\left(2\left(4\right)-1\right)\pi +\left(2\left(5\right)-1\right)\pi +\left(2\left(6\right)-1\right)\pi +\left(2\left(7\right)-1\right)\pi +\left(2\left(8\right)-1\right)\pi \\ =\left(2-1\right)\pi +\left(4-1\right)\pi +\left(6-1\right)\pi +\left(8-1\right)\pi +\left(10-1\right)\pi +\left(12-1\right)\pi +\left(14-1\right)\pi +\left(16-1\right)\pi \\ =\pi +3\pi +5\pi +7\pi +9\pi +11\pi +13\pi +15\pi \\ =64\pi \end{array}$

It can be observed that doubling the number of rings make the area of the target $4$ times bigger.

Thus, the value of the expression from part (b) when $n=1,2,4,8$ is $\pi ,4\pi ,16\pi ,64\pi$ and doubling the number of rings make the area of the target $4$ times bigger.