Chapter 6.FOM, Problem 1P

1-4 Modeling Periodic Data A set of data is given.

(a) Make a scatter plot of the data.

(b) Find a cosine function of the form $y=a\cos \left(\omega \left(t-c\right)\right)+b$ that models the data, as in Example 1.

(c) Graph the function you found in part (b) together with the scatter plot. How well does the curve fit the data?

(d) Use a graphing calculator to find the sine function that best fits the data, as in Example 2.

(e) Compare the functions you found in parts (b) and (d). [Use the reduction formula $\sin u=\cos (u-\pi /2).]$

t	y
0	2.1
2	1.1
4	$-0.8$
6	$-2.1$
8	$-1.3$
10	0.6
12	1.9
14	1.5

To determine

(a)

To find:

A scatter plot of the data.

Answer to Problem 1P

Solution:

Explanation of Solution

Calculation:

The scatter plot for the given data is given by

Final statement:

To determine

(b)

To find:

A cosine function of the form $y=a\cos \left(\omega \left(t-c\right)\right)+b$ that models the data.

Answer to Problem 1P

Solution:

$y=2.1\cos \left(0.52t\right).$

Explanation of Solution

Calculation:

Here, the maximum value is 2.1 and the minimum value is -2.1.

The maximum value occurs at time 0 and the minimum value occurs at time 6.

The vertical shift $b$ is the average of the maximum and minimum values.

$b=$ vertical shift

$b=\frac{1}{2}\times ($ maximum value + minimum value)

$b=\frac{1}{2}\left(2.1+\left(-2.1\right)\right)$

$b=\frac{1}{2}\left(2.1-2.1\right)$

$b=0.$

The amplitude $a$ is half of the difference between the maximum and minimum values.

$a=$ amplitude

$a=\frac{1}{2}\times ($ maximum value - minimum value)

$a=\frac{1}{2}\left(2.1-\left(-2.1\right)\right)$

$a=\frac{1}{2}\left(2.1+2.1\right)$

$a=\frac{1}{2}\left(4.2\right)$

$a=2.1.$

The time between consecutive maximum and minimum values is half of one period.

That is, $\frac{2\pi }{\omega }=$ period.

$\frac{2\pi }{\omega }=2\times ($ time of maximum value - time of minimum value)

$\frac{2\pi }{\omega }=2\left(0-6\right)$

$\frac{2\pi }{\omega }=-12$

$\omega =\frac{2\pi }{-12}$

$\omega =-\frac{\pi }{6}.$

Since the maximum value of the data occurs at $t=0,$ it represents a cosine curve.

There is no horizontal shift.

$c=0.$

Apply the values in the function $y=a\cos \left(\omega \left(t-c\right)\right)+b.$

$y=2.1\cos \left(-\frac{\pi }{6}\left(t-0\right)\right)+0$

$y=2.1\cos \left(-\frac{\pi }{6}t\right)$

Since cosine is an even function, $y=2.1\cos \left(\frac{\pi }{6}t\right).$

$y=2.1\cos \left(0.52t\right).$

Final statement:

$y=2.1\cos \left(0.52t\right).$

To determine

(c)

To find:

The graph of the function from part (b).

Answer to Problem 1P

Solution:

Explanation of Solution

Calculation:

The graph of the function $y=2.1\cos \left(0.52t\right)$ is given by

Final statement:

To determine

(d)

To find:

The sine function that fits the data.

Answer to Problem 1P

Solution:

$y=2.05\sin \left(0.50t+1.55\right)-0.01.$

Explanation of Solution

Calculation:

Using the given data and the SinReg command on the T1-83 calculator, we get a function of the form $y=a\sin \left(bt+c\right)+d$.

Here, $a=2.0487\dots .\approx 2.05$

$b=0.503079\dots \approx 0.50$

$c=1.55185\dots \approx 1.55$

$d=-0.008961\dots \approx -0.01$

Apply the values in the function $y=a\sin \left(bt+c\right)+d.$

$y=2.05\sin \left(0.50t+1.55\right)-0.01.$

Final statement:

$y=2.05\sin \left(0.50t+1.55\right)-0.01.$

To determine

(e)

To find:

The comparison between the functions from part (b) and (d).

Answer to Problem 1P

Solution:

$y=2.05\cos \left(0.50t-0.02\right)-0.01.$

It is same as (b), corrected to one decimal.

Explanation of Solution

Calculation:

From part (d), $y=2.05\sin \left(0.50t+1.55\right)-0.01.$

Apply the reduction formula $\sin x=\cos \left(x-\frac{\pi }{2}\right).$

So, $y=2.05\cos \left(0.50t+1.55-\frac{\pi }{2}\right)-0.01$

$y=2.05\cos \left(0.50t+1.55-1.57\right)-0.01$

$y=2.05\cos \left(0.50t-0.02\right)-0.01.$

It is same as (b), corrected to one decimal.

Final statement:

$y=2.05\cos \left(0.50t-0.02\right)-0.01.$

It is same as (b), corrected to one decimal.