Chapter 3.5, Problem 60E

a.

To determine

To calculate: The linear model $t_3$ and exponential model $t_4$ for the data.

  $\begin{array}{cc}\text{Speed, }s& \text{Time, }t\\ 30& 3.4\\ 40& 5.0\\ 50& 7.0\\ 60& 9.3\\ 70& 12.0\\ 80& 15.8\\ 90& 20.0\end{array}$

Answer to Problem 60E

The linear model $t_3$ is $t_3=0.273s-6.014$ and exponential model $t_4$ is $t_4=1.539{\left(1.03\right)}^s$ .

Explanation of Solution

Given information:

The time measured in seconds required by a car to attain a speed measured in miles from a starting point is provided below,

  $\begin{array}{cc}\text{Speed, }s& \text{Time, }t\\ 30& 3.4\\ 40& 5.0\\ 50& 7.0\\ 60& 9.3\\ 70& 12.0\\ 80& 15.8\\ 90& 20.0\end{array}$

Also, the above data can be modeled as,

  $\begin{array}{c}{t}_1=40.757+0.556s-15.817\ln s\\ t_2=1.2259+0.0023s^2\end{array}$

Calculation:

Consider the time measured in seconds required by a car to attain a speed measured in miles from a starting point is provided below,

  $\begin{array}{cc}\text{Speed, }s& \text{Time, }t\\ 30& 3.4\\ 40& 5.0\\ 50& 7.0\\ 60& 9.3\\ 70& 12.0\\ 80& 15.8\\ 90& 20.0\end{array}$

Follow the steps provided below to estimate the linear model.

Step 1: Press $\text{STAT}$ key.

Step 2: Press $\text{ENTER}$ key.

Step 3: Enter the column 1 of table provided above in $\text{L1}$ and press $\text{ENTER}$ key.

Step 4: Enter the column 2 of table provided above in $\text{L2}$ and press $\text{ENTER}$ key.

Step 5: Press $\text{2nd}$ key and $\text{MODE}$ key.

Step 6: Press $\text{STAT}$ key, select the $\text{CALC}$ menu.

Step 7: Select the fourth option that is $\text{LinReg}\left(ax+b\right)$ and press $\text{ENTER}$ key.

Step 8: Press $\text{2nd}$ key and press 1 to call list $\text{L1}$ and put comma.

Step 9: Press $\text{2nd}$ key and press 2 to call list $\text{L2}$ and press $\text{ENTER}$ key.

The result obtained on screen is provided below.

  

Therefore, linear model is,

  $\begin{array}{c}y=ax+b\\ t_3=0.273s-6.014\end{array}$

Follow the steps provided below to estimate the exponential model.

Step 1: Press $\text{STAT}$ key.

Step 2: Press $\text{ENTER}$ key.

Step 3: Enter the column 1 of table provided above in $\text{L1}$ and press $\text{ENTER}$ key.

Step 4: Enter the column 2 of table provided above in $\text{L2}$ and press $\text{ENTER}$ key.

Step 5: Press $\text{2nd}$ key and $\text{MODE}$ key.

Step 6: Press $\text{STAT}$ key, select the $\text{CALC}$ menu.

Step 7: Select the fourth option that is $\text{ExpReg}$ and press $\text{ENTER}$ key.

Step 8: Press $\text{2nd}$ key and press 1 to call list $\text{L1}$ and put comma.

Step 9: Press $\text{2nd}$ key and press 2 to call list $\text{L2}$ and press $\text{ENTER}$ key.

The result obtained on screen is provided below.

  

Therefore, exponential model is,

  $\begin{array}{c}y=a{b}^x\\ t_4=1.539{\left(1.03\right)}^s\end{array}$

b.

To determine

To graph: The model that represent the data.

  $\begin{array}{cc}\text{Speed, }s& \text{Time, }t\\ 30& 3.4\\ 40& 5.0\\ 50& 7.0\\ 60& 9.3\\ 70& 12.0\\ 80& 15.8\\ 90& 20.0\end{array}$

Explanation of Solution

Given information:

The time measured in seconds required by a car to attain a speed measured in miles from a starting point is provided below,

  $\begin{array}{cc}\text{Speed, }s& \text{Time, }t\\ 30& 3.4\\ 40& 5.0\\ 50& 7.0\\ 60& 9.3\\ 70& 12.0\\ 80& 15.8\\ 90& 20.0\end{array}$

Also, the above data can be modeled as,

  $\begin{array}{c}{t}_1=40.757+0.556s-15.817\ln s\\ t_2=1.2259+0.0023s^2\end{array}$

Graph:

Consider the time measured in seconds required by a car to attain a speed measured in miles from a starting point is provided below,

  $\begin{array}{cc}\text{Speed, }s& \text{Time, }t\\ 30& 3.4\\ 40& 5.0\\ 50& 7.0\\ 60& 9.3\\ 70& 12.0\\ 80& 15.8\\ 90& 20.0\end{array}$

Also, the above data can be modeled as,

  $\begin{array}{c}{t}_1=40.757+0.556s-15.817\ln s\\ t_2=1.2259+0.0023s^2\end{array}$

The linear model $t_3$ is $t_3=0.273s-6.014$ and exponential model $t_4$ is $t_4=1.539{\left(1.03\right)}^s$ .

Plot the data points and all the models in the coordinate plane.

  

The green curve line represents $t_3$ , orange curve represents $t_2$ , blue represents $t_1$ red curve represents $t_4$ black curve represents the data.

Interpretation:

All the models and the data points itself coincide with each other. All the models fits with the data.

c.

To determine

To calculate: The tabular estimation of the models for the data.

  $\begin{array}{cc}\text{Speed, }s& \text{Time, }t\\ 30& 3.4\\ 40& 5.0\\ 50& 7.0\\ 60& 9.3\\ 70& 12.0\\ 80& 15.8\\ 90& 20.0\end{array}$

Answer to Problem 60E

The estimation is provided below,

  $\begin{array}{cccccccc}s& 30& 40& 50& 60& 70& 80& 90\\ t_1& 3.6& 4.6& 6.7& 9.4& 12.5& 15.9& 19.6\\ t_2& 3.3& 4.9& 7.0& 9.5& 12.5& 15.9& 19.9\\ t_3& 2.2& 4.9& 7.6& 10.4& 13.1& 15.8& 18.5\\ t_4& 3.7& 5.0& 6.7& 9.0& 12.1& 16.3& 22.0\end{array}$

Explanation of Solution

Given information:

The time measured in seconds required by a car to attain a speed measured in miles from a starting point is provided below,

  $\begin{array}{cc}\text{Speed, }s& \text{Time, }t\\ 30& 3.4\\ 40& 5.0\\ 50& 7.0\\ 60& 9.3\\ 70& 12.0\\ 80& 15.8\\ 90& 20.0\end{array}$

Also, the above data can be modeled as,

  $\begin{array}{c}{t}_1=40.757+0.556s-15.817\ln s\\ t_2=1.2259+0.0023s^2\end{array}$

Calculation:

Consider the time measured in seconds required by a car to attain a speed measured in miles from a starting point is provided below,

  $\begin{array}{cc}\text{Speed, }s& \text{Time, }t\\ 30& 3.4\\ 40& 5.0\\ 50& 7.0\\ 60& 9.3\\ 70& 12.0\\ 80& 15.8\\ 90& 20.0\end{array}$

The linear model $t_3$ is $t_3=0.273s-6.014$ and exponential model $t_4$ is $t_4=1.539{\left(1.03\right)}^s$ .

For each value of s compute the value of t for different models.

  $\begin{array}{cccccccc}s& 30& 40& 50& 60& 70& 80& 90\\ t_1& 3.6& 4.6& 6.7& 9.4& 12.5& 15.9& 19.6\\ t_2& 3.3& 4.9& 7.0& 9.5& 12.5& 15.9& 19.9\\ t_3& 2.2& 4.9& 7.6& 10.4& 13.1& 15.8& 18.5\\ t_4& 3.7& 5.0& 6.7& 9.0& 12.1& 16.3& 22.0\end{array}$

d.

To determine

To calculate: The sum of absolute values of differences between data and estimated values provided by each model. Also explain the model that best fits the data based on sums.

Answer to Problem 60E

The sum is $t_1=2.0,t_2=2.1,t_3=5.6\text{ and }{t}_4=3.5$ . The model that best fits the data is $t_2$ .

Explanation of Solution

Given information:

The time measured in seconds required by a car to attain a speed measured in miles from a starting point is provided below,

  $\begin{array}{cc}\text{Speed, }s& \text{Time, }t\\ 30& 3.4\\ 40& 5.0\\ 50& 7.0\\ 60& 9.3\\ 70& 12.0\\ 80& 15.8\\ 90& 20.0\end{array}$

Also, the above data can be modeled as,

  $\begin{array}{c}{t}_1=40.757+0.556s-15.817\ln s\\ t_2=1.2259+0.0023s^2\end{array}$

Calculation:

Consider the time measured in seconds required by a car to attain a speed measured in miles from a starting point is provided below,

  $\begin{array}{cc}\text{Speed, }s& \text{Time, }t\\ 30& 3.4\\ 40& 5.0\\ 50& 7.0\\ 60& 9.3\\ 70& 12.0\\ 80& 15.8\\ 90& 20.0\end{array}$

Also, the above data can be modeled as,

  $\begin{array}{c}{t}_1=40.757+0.556s-15.817\ln s\\ t_2=1.2259+0.0023s^2\end{array}$

The linear model $t_3$ is $t_3=0.273s-6.014$ and exponential model $t_4$ is $t_4=1.539{\left(1.03\right)}^s$ .

For each value of s compute the value of t for different models.

  $\begin{array}{cccccccc}s& 30& 40& 50& 60& 70& 80& 90\\ t_1& 3.6& 4.6& 6.7& 9.4& 12.5& 15.9& 19.6\\ t_2& 3.3& 4.9& 7.0& 9.5& 12.5& 15.9& 19.9\\ t_3& 2.2& 4.9& 7.6& 10.4& 13.1& 15.8& 18.5\\ t_4& 3.7& 5.0& 6.7& 9.0& 12.1& 16.3& 22.0\end{array}$

Now, sum of absolute values of differences between data and estimated values provided by each model is,

For $t_1$ ,

  $\left|3.6-3.4\right|+\left|4.6-5.0\right|+\left|6.7-7.0\right|+\left|9.4-9.3\right|+\left|12.5-12.0\right|+\left|15.9-15.8\right|+\left|19.6-20.0\right|=2.0$

For $t_2$ ,

  $\left|3.3-3.4\right|+\left|4.9-5.0\right|+\left|7.0-7.0\right|+\left|9.5-9.3\right|+\left|12.5-12.0\right|+\left|15.9-15.8\right|+\left|19.9-20.0\right|=2.1$

For $t_3$ ,

  $\left|2.2-3.4\right|+\left|4.9-5.0\right|+\left|7.6-7.0\right|+\left|10.4-9.3\right|+\left|13.1-12.0\right|+\left|15.8-15.8\right|+\left|18.5-20.0\right|=5.6$

For $t_4$ ,

  $\left|3.7-3.4\right|+\left|5.0-5.0\right|+\left|6.7-7.0\right|+\left|9.0-9.3\right|+\left|12.1-12.0\right|+\left|16.3-15.8\right|+\left|22.0-20.0\right|=3.5$

The model that best fits the data is the one which has least sum of the absolute values of the differences between the estimated and actual data. This is represented by model $t_2=1.2259+0.0023s^2$ .

Thus, the sum is $t_1=2.0,t_2=2.1,t_3=5.6\text{ and }{t}_4=3.5$ . The model that best fits the data is $t_2$ .