Chapter 4.5, Problem 66E

Solution

a.

To prove: $\frac{1}{y}=\frac{1}{a}\sin \left(bx+\frac{\pi }{2}\right)$ .

Given information:

  $\begin{array}{c}\frac{1}{y}=\frac{1}{a\sec \left(bx\right)}\\ =\frac{1}{a}\cos \left(bx\right)\\ \cos \left(bx\right)=\sin \left(bx+\frac{\pi }{2}\right)\end{array}$

Calculation:

From the given facts,

  $\begin{array}{c}\frac{1}{y}=\frac{1}{a}\cos \left(bx\right)\mathrm{......}\text{(1)}\\ \cos \left(bx\right)=\sin \left(bx+\frac{\pi }{2}\right)\mathrm{......}\text{(2)}\end{array}$

Substitute the equation (2) in (1) to get,

  $\frac{1}{y}=\frac{1}{a}\sin \left(bx+\frac{\pi }{2}\right)$

Therefore, $\frac{1}{y}=\frac{1}{a}\sin \left(bx+\frac{\pi }{2}\right)$ has been proved.

b.

To find: The regression equation.

The regression equation is $\frac{1}{y}=0.2\sin \left(\frac{1}{6}x+\frac{\pi }{2}\right)$ .

Given information:

  $\begin{array}{c}\frac{1}{y}=\frac{1}{a\sec \left(bx\right)}\\ =\frac{1}{a}\cos \left(bx\right)\\ \cos \left(bx\right)=\sin \left(bx+\frac{\pi }{2}\right)\end{array}$

Calculation:

Now, entering the values of $x$ -values on $L_1$ and $y$ -values on $L_2$ to get,

$L_1$	$L_2$	$L_3$
$1$	$5.0703$	$0.19723$
$2$	$5.2912$	$0.18899$
$3$	$5.6975$	$0.17552$
$4$	$6.3622$	$0.15718$
$5$	$7.4359$	$0.13448$
$6$	$9.2541$	$0.10806$
$7$	$12.716$	$0.07864$
$L_3=\mathrm{0.1972269885....}$

The sine regression is,

  $\begin{array}{c}y=a\ast \sin \left(bx+c\right)+d\\ a=0.1998918699\\ \approx 0.2\\ b=0.1667309045\\ \approx \frac{1}{6}\\ c=1.570686742\\ \approx \frac{\pi }{2}\\ d=1.037504E^{-4}\\ \approx 0\end{array}$

Substituting the above values in $\frac{1}{y}=\frac{1}{a}\sin \left(bx+\frac{\pi }{2}\right)$ to get,

  $\frac{1}{y}=0.2\sin \left(\frac{1}{6}x+\frac{\pi }{2}\right)$

Therefore, the regression equation is $\frac{1}{y}=0.2\sin \left(\frac{1}{6}x+\frac{\pi }{2}\right)$ .

c.

To find: The values of $a$ and $b$ .

The values of $a=5$ and $b=\frac{1}{6}$ .

Given information:

  $\begin{array}{c}\frac{1}{y}=\frac{1}{a\sec \left(bx\right)}\\ =\frac{1}{a}\cos \left(bx\right)\\ \cos \left(bx\right)=\sin \left(bx+\frac{\pi }{2}\right)\end{array}$

Calculation:

From part (b) the obtained equation is,

  $\frac{1}{y}=0.2\sin \left(\frac{1}{6}x+\frac{\pi }{2}\right)$

Comparing the above equation with

  $\frac{1}{y}=\frac{1}{a}\sin \left(bx+\frac{\pi }{2}\right)$

Thus,

  $\begin{array}{c}\frac{1}{a}=0.2\\ \frac{1}{a}=\frac{1}{5}\\ a=5\\ b=\frac{1}{6}\end{array}$

Therefore, the values of $a=5$ and $b=\frac{1}{6}$ .

d.

To find: The secant model and to check whether the curve fits the scatter plot.

The secant model is $y=5\sec \left(\frac{1}{6}x\right)$ .

The curve fits in the scatter plot $\left(L_1,L_2\right)$ .

Given information:

  $\begin{array}{c}\frac{1}{y}=\frac{1}{a\sec \left(bx\right)}\\ =\frac{1}{a}\cos \left(bx\right)\\ \cos \left(bx\right)=\sin \left(bx+\frac{\pi }{2}\right)\end{array}$ .

Calculation:

From the part (c), using the values of $a$ and $b$

The secant model is,

  $y=5\sec \left(\frac{1}{6}x\right)$

Graph:

Plotting the scatter value by substituting the values in $\frac{5}{\cos \left(\frac{1}{6}x\right)}$ .

  

Interpretation:

From the graph, it is found that the curve fits the $\left(L_1,L_2\right)$ scatter plot.