Chapter 11.6, Problem 9P

(a)

To determine

To find: the growth of vine during its 26^th weeks.

Answer to Problem 9P

  $0.188\text{ inch}$ .

Explanation of Solution

Given:

Vine growth during the week after it emerges from soil is $50\text{inch}$ .

During each succeeding week it grows $\frac{4}{5}$ .

Concept used:

The summation of infinite geometric series between $-1<r<1$ .

  $S=\frac{a_1}{1-r}$ .

Sigma notation to represent an infinite series.

  ${{\displaystyle {\sum }_{k=1}^{\infty }{a}_1\left(\frac{1}{2}\right)}}^{k-1}$ .

Here $\frac{1}{2}$ is common ratio and $k$ number of terms.

Calculation:

The growth in week $n$ can be written recursively as:

  $\begin{array}{l}g\left(0\right)=50.\\ g\left(n+1\right)=\frac{4}{5}g\left(n\right).\end{array}$

This is a geometrical sequence with ratio $\frac{4}{5}$ .

  $g\left(n\right)=50.{\left(\frac{4}{5}\right)}^n$ .

The growth during the $26th$ week is:

  $g\left(25\right)={\mathrm{50.0.8}}^{25}\simeq 0.188$ .

Hence, $0.188\text{ inch}$ .

(b)

To determine

To find: the total growth for a vine which grows during its 26^th week.

Answer to Problem 9P

  $249.244\text{ inch}$ .

Explanation of Solution

Given:

Vine growth during the week after it emerges from soil is $50\text{inch}$ .

During each succeeding week it grows $\frac{4}{5}$ .

Concept used:

The summation of infinite geometric series between $-1<r<1$ .

  $S=\frac{a_1}{1-r}$ .

Sigma notation to represent an infinite series.

  ${{\displaystyle {\sum }_{k=1}^{\infty }{a}_1\left(\frac{1}{2}\right)}}^{k-1}$ .

Here $\frac{1}{2}$ is common ratio and $k$ number of terms.

Calculation:

The growth in week $n$ can be written recursively as:

  $\begin{array}{l}g\left(0\right)=50.\\ g\left(n+1\right)=\frac{4}{5}g\left(n\right).\end{array}$

This is a geometrical sequence with ratio $\frac{4}{5}$ .

  $g\left(n\right)=50.{\left(\frac{4}{5}\right)}^n$ .

The total growth during the $26th$ week is:

  ${{\displaystyle {\sum }_{n=0}^{25}50\times 0.8}}^n=50\frac{1-{0.8}^{26}}{1-0.8}\simeq \mathrm{249.244.}$ .

Hence, $249.244\text{ inch}$ .

(c)

To determine

To find: the total growth of vine assuming that the vine grow forever.

Answer to Problem 9P

  $250\text{ inch}$ .

Explanation of Solution

Given:

Vine growth during the week after it emerges from soil is $50\text{inch}$ .

During each succeeding week it grows $\frac{4}{5}$ .

Concept used:

The summation of infinite geometric series between $-1<r<1$ .

  $S=\frac{a_1}{1-r}$ .

Sigma notation to represent an infinite series.

  ${{\displaystyle {\sum }_{k=1}^{\infty }{a}_1\left(\frac{1}{2}\right)}}^{k-1}$ .

Here $\frac{1}{2}$ is common ratio and $k$ number of terms.

Calculation:

The growth in week $n$ can be written recursively as:

  $\begin{array}{l}g\left(0\right)=50.\\ g\left(n+1\right)=\frac{4}{5}g\left(n\right).\end{array}$

This is a geometrical sequence with ratio $\frac{4}{5}$ .

  $g\left(n\right)=50.{\left(\frac{4}{5}\right)}^n$ .

The total growth in the series:

  ${\displaystyle {\sum }_0^{\infty }g\left(n\right)=50{\displaystyle {\sum }_0^{\infty }{0.8}^n}}=\frac{50}{1-0.8}\simeq 250.$ .

Hence, $250\text{ inch}$ .