Chapter 1.7, Problem 28E

a.

To determine

To find: the type of function represented the degree days.

Answer to Problem 28E

Absolute value function

Explanation of Solution

Given information:

The number of degree days = $d\left(t\right)$

Mean temperature = $t°\text{F}$

Number of degree days is the difference between 65 and the mean temperature.

If $t°\text{F}>65°\text{F}$ , cooling degree days are recorded.

If $t°\text{F}<65°\text{F}$ , heating degree days are recorded.

Calculation:

Since the table shows different linear rates, the function of degree days due to mean temperature should be a absolute value function.

Thus, the type of function for the given situation is absolute value function.

b.

To determine

To find: the function to model $d\left(t\right)$ for $t°\text{F}$ .

Answer to Problem 28E

  $d\left(t\right)=\left|65-t\right|$

Explanation of Solution

Given information:

The number of degree days = $d\left(t\right)$

Mean temperature = $t°\text{F}$

Number of degree days is the difference between 65 and the mean temperature.

If $t°\text{F}>65°\text{F}$ , cooling degree days are recorded.

If $t°\text{F}<65°\text{F}$ , heating degree days are recorded.

Calculation:

Since the number of degree days is the difference between 65 and the mean temperature. The function of $d\left(t\right)$ for $t°\text{F}$ is as follows:

  $d\left(t\right)=\left|65-t\right|$

c.

To determine

To graph: the function.

Answer to Problem 28E

  

Explanation of Solution

Given information:

The number of degree days = $d\left(t\right)$

Mean temperature = $t°\text{F}$

  $d\left(t\right)=\left|65-t\right|$

Number of degree days is the difference between 65 and the mean temperature.

If $t°\text{F}>65°\text{F}$ , cooling degree days are recorded.

If $t°\text{F}<65°\text{F}$ , heating degree days are recorded.

Calculation:

The table of values given is as follows:

$t$	$d\left(t\right)$
25	40
35	30
45	20
55	10
65	0
75	10
85	20
95	30

The graph for the above function is as follows:

  

d.

To determine

To find: $d\left(t\right)$ that is recorded for a day with high temperature and low temperature, and mention whether they are heating or cooling degree days.

Answer to Problem 28E

  $\begin{array}{l}d\left(63°\text{F}\right)=2°\text{F}\\ d\left(28°\text{F}\right)=37°\text{F}\end{array}$

Heating days

Explanation of Solution

Given information:

The number of degree days = $d\left(t\right)$

Mean temperature = $t°\text{F}$

  $d\left(t\right)=\left|65-t\right|$

Number of degree days is the difference between 65 and the mean temperature.

If $t°\text{F}>65°\text{F}$ , cooling degree days are recorded.

If $t°\text{F}<65°\text{F}$ , heating degree days are recorded.

High temperature = $63°\text{F}$

Low temperature = $28°\text{F}$

Calculation:

Substitute the value of high temperature in the function to evaluate $d\left(t\right)$ as follows:

  $\begin{array}{c}d\left(t\right)=\left|65-t\right|\\ =\left|65-63\right|\\ =\left|2\right|\\ =2\end{array}$

Thus, $d\left(t\right)$ that is recorded for a day with high temperature is $2°\text{F}$ .

Substitute the value of low temperature in the function to evaluate $d\left(t\right)$ as follows:

  $\begin{array}{c}d\left(t\right)=\left|65-t\right|\\ =\left|65-28\right|\\ =\left|37\right|\\ =37\end{array}$

Thus, $d\left(t\right)$ that is recorded for a day with low temperature is $37°\text{F}$ .

Since $t°\text{F}<65°\text{F}$ , heating degree days are recorded. Thus, the degree days for given temperatures is heating degree days.