Chapter 1.7, Problem 19E

(a)

To determine

To graph: The scatter plot of the data.

Explanation of Solution

Given information:

Hooke’s law states that the force $F$ required to compress or stretch a spring(within elastic limits) is proportional to the distance $d$ that the spring is compressed or stretched from its original length. That is, $F=kd$ , where $k$ is the measure of the stiffness of the spring and is called the spring constant. The table shows the elongation $d$ in centimeters of spring when a force of $F$ kilograms is applied.

Force, $F$	Elongation, $d$
$20$	$1.4$
$40$	$2.5$
$60$	$4.0$
$80$	$5.3$
$100$	$6.6$

Graph:

To graph the points on scatter plot, follow the steps using graphing calculator.

First press the $\boxed{2\text{nd}}$ key and then $\boxed{\text{Y}=}$ and make sure that Plot $1$ is ON and choose the scatter plot icon in type.

Go to ${\text{Y}}_1$ and then press the key $\boxed{\text{STAT}}$ and then in edit enter the data in ${\text{L}}_1$ and ${\text{L}}_2$ .

Now, press the $\boxed{\text{ZOOM}}\boxed{9}$ key and $\boxed{\text{TRACE}}$ key to produce the graph in standard window as shown in Figure (1).

  

Figure (1)

(b)

To determine

To find: The equation of line that seems to best fit for the data.

Answer to Problem 19E

The equation of line that seems to best fit for the data is equal to $d=0.055F+0.3$ .

Explanation of Solution

Given information:

Hooke’s law states that the force $F$ required to compress or stretch a spring(within elastic limits) is proportional to the distance $d$ that the spring is compressed or stretched from its original length. That is, $F=kd$ , where $k$ is the measure of the stiffness of the spring and is called the spring constant. The table shows the elongation $d$ in centimeters of spring when a force of $F$ kilograms is applied.

Force, $F$	Elongation, $d$
$20$	$1.4$
$40$	$2.5$
$60$	$4.0$
$80$	$5.3$
$100$	$6.6$

Calculation:

Assume that the line is equal to $d=aF+b$ .

As given the value of $d$ is $1.4$ for $F=20$ and the value of $d$ is $2.5$ for $F=40$ .

Substitute $20$ for $F$ and $1.4$ for $d$ in the general equation.

  $\begin{array}{c}d=aF+b\\ 1.4=a\left(20\right)+b\\ 20a+b=1.4\end{array}$

Substitute $40$ for $F$ and $2.5$ for $d$ in the general equation.

  $\begin{array}{c}d=aF+b\\ 2.5=a\left(40\right)+b\\ 40a+b=2.5\end{array}$

Subtract both the equations.

  $\begin{array}{c}40a+b-\left(20a+b\right)=2.5-1.4\\ 20a=1.1\\ a=\frac{1.1}{20}\\ a=0.055\end{array}$

Substitute $1.1$ for $20a$ in the first equation.

  $\begin{array}{c}20a+b=1.4\\ 1.1+b=1.4\\ b=1.4-1.1\\ b=0.3\end{array}$

Therefore, the equation of line that seems to best fit for the data is equal to $d=0.055F+0.3$ .

(c)

To determine

To find: The linear model for the data with the help of regression of graphing utility.

Answer to Problem 19E

The linear model for the data is equal to $d=0.066F$ .

Explanation of Solution

Given information:

Hooke’s law states that the force $F$ required to compress or stretch a spring(within elastic limits) is proportional to the distance $d$ that the spring is compressed or stretched from its original length. That is, $F=kd$ , where $k$ is the measure of the stiffness of the spring and is called the spring constant. The table shows the elongation $d$ in centimeters of spring when a force of $F$ kilograms is applied.

Force, $F$	Elongation, $d$
$20$	$1.4$
$40$	$2.5$
$60$	$4.0$
$80$	$5.3$
$100$	$6.6$

Calculation:

For the regression linear model of data follow the steps below:

First press the $\boxed{2\text{nd}}$ key and then $\boxed{\text{Y}=}$ and make sure that Plot $1$ is ON and choose the scatter plot icon in type.

Go to ${\text{Y}}_1$ and then press the key $\boxed{\text{STAT}}$ and then in edit enter the data in ${\text{L}}_1$ and ${\text{L}}_2$ .

Now enter the keystrokes $\boxed{\text{STAT}}\boxed{\to }\boxed{4}$ and press the $\boxed{\text{ENTER}}$ key for the regression.

  

Figure(1)

Therefore, the linear model for the data is equal to $d=0.066F$ .

(d)

To determine

To find: The elongation of the spring when a force of $55$ kilograms is applied with the help of model in part(c).

Answer to Problem 19E

The elongation of the spring when a force of $55$ kilograms is applied is equal to $3.63\text{cm}$ .

Explanation of Solution

Given information:

Hooke’s law states that the force $F$ required to compress or stretch a spring(within elastic limits) is proportional to the distance $d$ that the spring is compressed or stretched from its original length. That is, $F=kd$ , where $k$ is the measure of the stiffness of the spring and is called the spring constant. The table shows the elongation $d$ in centimeters of spring when a force of $F$ kilograms is applied.

Force, $F$	Elongation, $d$
$20$	$1.4$
$40$	$2.5$
$60$	$4.0$
$80$	$5.3$
$100$	$6.6$

Calculation:

As found in part(c), the model for the best fit of data is equal to $d=0.066F$ .

Substitute $55$ for $F$ in the model.

  $\begin{array}{l}d=0.066F\\ d=0.066\left(55\right)\\ d=3.63\end{array}$

Therefore, the elongation of the spring when a force of $55$ kilograms is applied is equal to $3.63\text{cm}$ .