Chapter 12.4, Problem 12CFU

(a)

To determine

To find: the distance of spill spreads after $10$ years.

Answer to Problem 12CFU

The chemical had spread $15\text{m}$ more.

Explanation of Solution

Given:

The chemical has spread after two years is:

  $1500\text{m}$ .

The chemical had spread more is:

  $900\text{m}$ .

The chemical had spread by the end of the third year is:

  $540\text{m}$ .

Concept used:

If there is common ratio then the series is geometric series.

Common ratio is:

  $\begin{array}{l}r=\frac{t_2}{t_1}=\frac{t_3}{t_2}=\frac{t_4}{t_3}=k.\\ t_n=a{r}^{n-1}.\\ s_n=\frac{r\left(1-r^n\right)}{1-r}.\end{array}$

Calculation:

Consider the chemical has spread after two years is:

  $1500\text{m}$ .

The chemical had spread more is:

  $900\text{m}$ .

The chemical had spread by the end of the third year is:

  $540\text{m}$ .

The sum of the series from the given data is:

  $1500+900+540+\mathrm{....}$

This can be written as:

  $1500\left(1+\frac{3}{5}+\frac{9}{25}+\mathrm{....}\right)$

Here $r=\frac{3}{5}$ .

The source after $10$ year is:

  $\begin{array}{l}{t}_{10}={\left(\frac{3}{5}\right)}^9\times 1500.\\ \Rightarrow {\left(0.6\right)}^9\times 1500.\\ \Rightarrow \mathrm{15.1165.}\end{array}$

Hence, the chemical had spread $15\text{m}$ more.

(b)

To determine

To find: whether the spill ever reach the grounds of a school located from the source.

Answer to Problem 12CFU

The spill will never reach the grounds of a school located at $4000\text{m}$ away from the source.

Explanation of Solution

Given:

The chemical has spread after two years is:

  $1500\text{m}$ .

The chemical had spread more is:

  $900\text{m}$ .

The chemical had spread by the end of the third year is:

  $540\text{m}$ .

Concept used:

If there is common ratio then the series is geometric series.

Common ratio is:

  $\begin{array}{l}r=\frac{t_2}{t_1}=\frac{t_3}{t_2}=\frac{t_4}{t_3}=k.\\ t_n=a{r}^{n-1}.\\ s_n=\frac{r\left(1-r^n\right)}{1-r}.\end{array}$

Calculation:

Using $\frac{a}{1-r}$

The expression can be solved as:

  $\begin{array}{l}\frac{1500}{1-r}=\frac{1500}{1-\frac{3}{5}}.\\ \Rightarrow \frac{1500}{\frac{2}{5}}=\frac{7550}{2}.\\ \Rightarrow 3750<4000.\end{array}$

Hence, the spill will never reach the grounds of a school located at $4000\text{m}$ away from the source.