Chapter 8.3, Problem 107E

(a)

To determine

To find: a formula for the $n$ th term of the geometric sequence that gives the temperature of the water $n$ hours after it is placed in the freezer.

Answer to Problem 107E

The $n$ th term of the above geometric sequence is given by

  $T_n=70{\left(0.8\right)}^n$

Explanation of Solution

Given information:

The temperature of water in an ice tray is $70°\text{F}$ when is is placed in a freezer. Its temperature $n$ hours after being placed in the freezer is 20% less than 1 hour earlier.

Calculation:

Now determine a formula for the $n$ th term of the geometric sequence that gives the temperature of the water $n$ hours after being placed in the freezer as shown below:

Here, the sequence of temperature of the water follows a geometric sequence as shown below:

  $\begin{array}{c}70°\text{F}{\left(1-20\%\right)}^1,70°\text{F}{\left(1-20\%\right)}^2,70°\text{F}{\left(1-20\%\right)}^3,\mathrm{...},70°\text{F}{\left(1-20\%\right)}^n,\mathrm{...}\\ =70°\text{F}{\left(0.8\right)}^1,70°\text{F}{\left(0.8\right)}^2,70°\text{F}{\left(0.8\right)}^3,\mathrm{...},70°\text{F}{\left(0.8\right)}^n,\mathrm{...}\\ =70°\text{F}{\left(0.8\right)}^1,70°\text{F}{\left(0.8\right)}^2,\mathrm{...70}°\text{F}{\left(0.8\right)}^n,\mathrm{...}\end{array}$

The $n$ th term of the above geometric sequence is given by

  $T_n=70{\left(0.8\right)}^n$

(b)

To determine

To find: the temperature of the water 6 hours and 12 hours after it is placed in the freezer.

Answer to Problem 107E

The temperature of the water 6 hours and 12 hours after it is placed in the freezer are $18.4°\text{F}$ and $4.8°\text{F}$ .

Explanation of Solution

Given information:

The temperature of water in an ice tray is $70°\text{F}$ when is is placed in a freezer. Its temperature $n$ hours after being placed in the freezer is 20% less than 1 hour earlier.

Calculation:

Now determine the temperature of the water 6 hours and 12 hours after it is placed in the freezer are obtained as shown below:

  $\begin{array}{c}{T}_n=70{\left(0.8\right)}^n\\ T_6=70{\left(0.8\right)}^6\\ =18.35008\\ =18.4°\text{F}\end{array}$

Finding the temperature of water 12 hours after it is placed in the freezer as shown belo:

  $\begin{array}{c}{T}_n=70{\left(0.8\right)}^n\\ T_{12}=70{\left(0.8\right)}^{12}\\ =1.81036337152\\ =4.8°\text{F}\end{array}$

Thus, the temperature of the water 6 hours and 12 hours after it is placed in the freezer are $18.4°\text{F}$ and $4.8°\text{F}$ .

(c)

To determine

To approximate: the time required for the water to freeze using graphing utility.

Answer to Problem 107E

The approximate time required for the water to freeze by using a graphing utility is 3.5 hours.

Explanation of Solution

Given information:

The temperature of water in an ice tray is $70°\text{F}$ when is is placed in a freezer. Its temperature $n$ hours after being placed in the freezer is 20% less than 1 hour earlier.

Calculation:

The freezing point of water is $32°\text{F}$ the approximate value of the ccorresponding time is obtained as shown below:

  $\begin{array}{c}{T}_n=70{\left(0.8\right)}^n\\ 32=70{\left(0.8\right)}^n\\ 16=35{\left(0.8\right)}^n\\ {\left(0.8\right)}^n=\frac{16}{35}\end{array}$

Continue simplification

  $\begin{array}{c}{\log }_{0.8}{\left(0.8\right)}^n={\log }_{0.8}\left(\frac{16}{35}\right)\\ n={\log }_{0.8}\left(\frac{16}{35}\right)\\ =\frac{{\log }_{10}\left(\frac{16}{35}\right)}{{\log }_{10}\left(0.8\right)}\\ =3.507873\\ \approx 3.5\text{hours}\end{array}$

The corresponding graph is given by

  

Thus, the approximate time required for the water to freeze by using a graphing utility is 3.5 hours.