Chapter 12, Problem 12PS

a.

To determine

To find: a quadratic model for the given data using the regression feature of a graphing utility of the form $T_1=a{t}^2+bt+c$ for the data.

Answer to Problem 12PS

Quadratic model is $T_1=-0.0030279t^2-0.677237t+26.5636$ .

Explanation of Solution

Given information: The heat exchange of a heating system has a probe attached to it. The probe records the temperature T (in degrees Celsius) t seconds after the furnace is started. The table shows the results for the first 2 minutes.

t	T
0	25.2 ${}^{\circ }$
15	36.9 ${}^{\circ }$
30	45.5 ${}^{\circ }$
45	51.4 ${}^{\circ }$
60	56.0 ${}^{\circ }$
75	59.6 ${}^{\circ }$
90	62.0 ${}^{\circ }$
105	64.0 ${}^{\circ }$
120	65.2 ${}^{\circ }$

Calculation:

Using the graphing utility on above table data to find the model:

  $\begin{array}{c}\text{QuadReg}\\ T_1=a{t}^2+bt+c.\\ a=-\mathrm{0.0030279.}\\ b=\mathrm{0.677237.}\\ c=\mathrm{26.5636.}\end{array}$

  $T_1=-0.0030279t^2-0.677237t+\mathrm{26.5636.}$

b.

To determine

To graph: the model $T_1$ using graphing utility and how well does the model fit the data.

Answer to Problem 12PS

The curve fits the data almost correctly .But all points are not exactly same as the data.

Explanation of Solution

Given information: The heat exchange of a heating system has a probe attached to it. The probe records the temperature T (in degrees Celsius) t seconds after the furnace is started. The table shows the results for the first 2 minutes.

t	T
0	25.2 ${}^{\circ }$
15	36.9 ${}^{\circ }$
30	45.5 ${}^{\circ }$
45	51.4 ${}^{\circ }$
60	56.0 ${}^{\circ }$
75	59.6 ${}^{\circ }$
90	62.0 ${}^{\circ }$
105	64.0 ${}^{\circ }$
120	65.2 ${}^{\circ }$

Calculation:

Quadratic model is $T_1=-0.0030279t^2-0.677237t+26.5636$ .

The graph of this model using graphing utility is shown below.

  

The curve fits the data almost correctly .But all points are not exactly same as the data.

c.

To determine

To graph: the model $T_2$ with the original data and how well does the model fit the data.

Answer to Problem 12PS

Here the curve fits the actual data almost at every point.

Explanation of Solution

Given information: The heat exchange of a heating system has a probe attached to it. The probe records the temperature T (in degrees Celsius) t seconds after the furnace is started. The table shows the results for the first 2 minutes.

t	T
0	25.2 ${}^{\circ }$
15	36.9 ${}^{\circ }$
30	45.5 ${}^{\circ }$
45	51.4 ${}^{\circ }$
60	56.0 ${}^{\circ }$
75	59.6 ${}^{\circ }$
90	62.0 ${}^{\circ }$
105	64.0 ${}^{\circ }$
120	65.2 ${}^{\circ }$

A ration model for the data is given by,

  $T_2=\frac{86t+1451}{t+58}.$

Calculation:

The graph of this $T_2$ model using graphing utility is shown below

.

  

Here the curve fits the actual data almost at every point.

d.

To determine

To evaluate: $T_1(0)and{T}_2(0).$

Answer to Problem 12PS

  $T_1(0)=\mathrm{26.5636.}$

  $T_2=\mathrm{25.017.}$

Explanation of Solution

Given information The heat exchange of a heating system has a probe attached to it. The probe records the temperature T (in degrees Celsius) t seconds after the furnace is started. The table shows the results for the first 2 minutes.

t	T
0	25.2 ${}^{\circ }$
15	36.9 ${}^{\circ }$
30	45.5 ${}^{\circ }$
45	51.4 ${}^{\circ }$
60	56.0 ${}^{\circ }$
75	59.6 ${}^{\circ }$
90	62.0 ${}^{\circ }$
105	64.0 ${}^{\circ }$
120	65.2 ${}^{\circ }$

Calculation:

  $\begin{array}{l}{T}_1=-0.0030279t^2-0.677237t+\mathrm{26.5636.}\\ T_1(0)=-0.0030279\times {(}^0-0.677237\times (0)+\mathrm{26.5636.}\\ T_1(0)=\mathrm{26.5636.}\end{array}$

  $\begin{array}{l}{T}_2=\frac{86t+1451}{t+58}.\\ T_2(0)=\frac{86\times 0+1451}{0+58}=\mathrm{25.017.}\\ T_2=\mathrm{25.017.}\end{array}$

e.

To determine

To find : $\underset{t\to \infty }{\lim }{T}_2$ .Verify results using the graph in part (c ).

Answer to Problem 12PS

  $\underset{t\to \infty }{\lim }{T}_2=$ 86.

Explanation of Solution

Given information The heat exchange of a heating system has a probe attached to it. The probe records the temperature T (in degrees Celsius) t seconds after the furnace is started. The table shows the results for the first 2 minutes.

t	T
0	25.2 ${}^{\circ }$
15	36.9 ${}^{\circ }$
30	45.5 ${}^{\circ }$
45	51.4 ${}^{\circ }$
60	56.0 ${}^{\circ }$
75	59.6 ${}^{\circ }$
90	62.0 ${}^{\circ }$
105	64.0 ${}^{\circ }$
120	65.2 ${}^{\circ }$

Calculation:

  $\begin{array}{c}\underset{t\to \infty }{\lim }{T}_2=\underset{t\to \infty }{\lim }\frac{86t+1451}{t+58}=\underset{t\to \infty }{\lim }\frac{86+\frac{1451}{t}}{1+\frac{58}{t}}.\\ \text{as}t\to \infty ,\frac{1}{t}\to 0\text{thus}\\ \underset{t\to \infty }{\lim }\frac{86+\frac{1451}{t}}{1+\frac{58}{t}}=\frac{86}{1}=86.\\ \underset{t\to \infty }{\lim }{T}_2=86.\end{array}$

  

f.

To determine

To interpret: the result of part (c) in context of the problem and is it possible to perform this type of analysis using $T_1$ .

Answer to Problem 12PS

The result says that after a long period of time the temperature will reach a saturated value , that is 86 ${}^{\circ }$ C . By performing analysis with $T_1$ , $\underset{t\to \infty }{\lim }a{t}^2+bt+c$ does not exist for quadratic function. Thus not possible

Explanation of Solution

Given information The heat exchange of a heating system has a probe attached to it. The probe records the temperature T (in degrees Celsius) t seconds after the furnace is started. The table shows the results for the first 2 minutes.

t	T
0	25.2 ${}^{\circ }$
15	36.9 ${}^{\circ }$
30	45.5 ${}^{\circ }$
45	51.4 ${}^{\circ }$
60	56.0 ${}^{\circ }$
75	59.6 ${}^{\circ }$
90	62.0 ${}^{\circ }$
105	64.0 ${}^{\circ }$
120	65.2 ${}^{\circ }$

Calculation:

The result says that after a long period of time the temperature will reach a saturated value, that is 86 ${}^{\circ }$ C . By performing analysis with $T_1$ , $\underset{t\to \infty }{\lim }a{t}^2+bt+c$ does not exist for quadratic function. Thus not possible.