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Algebra and Trigonometry (MindTap ...

4th Edition
James Stewart + 2 others
ISBN: 9781305071742

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Algebra and Trigonometry (MindTap ...

4th Edition
James Stewart + 2 others
ISBN: 9781305071742
Textbook Problem
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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 ω t - c + 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 - π / 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

Solution:

Algebra and Trigonometry (MindTap Course List), Chapter 6.FOM, Problem 1P , additional homework tip  1

Explanation

Calculation:

The scatter plot for the given data is given by

Algebra and Trigonometry (MindTap Course List), Chapter 6.FOM, Problem 1P , additional homework tip  2

Final statement:

Algebra and Trigonometry (MindTap Course List), Chapter 6.FOM, Problem 1P , additional homework tip  3

To determine

(b)

To find:

A cosine function of the form y=acosωt-c+b that models the data.

Answer

Solution:

y=2.1cos0.52t.

Explanation

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=12×( maximum value + minimum value)

b=122.1+-2.1

b=122.1-2.1

b=0.

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

a= amplitude

a=12×( maximum value - minimum value)

a=122.1--2.1

a=122.1+2.1

a=124.2

a=2.1.

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

That is, 2πω= period.

2πω=2×( time of maximum value - time of minimum value)

2πω=20-6

2πω=-12

ω=2π-12

ω=-π6.

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=acosωt-c+b.

y=2.1cos-π6t-0+0

y=2.1cos-π6t

Since cosine is an even function, y=2.1cosπ6t.

y=2.1cos0.52t.

Final statement:

y=2.1cos0.52t.

To determine

(c)

To find:

The graph of the function from part (b).

Answer

Solution:

Algebra and Trigonometry (MindTap Course List), Chapter 6.FOM, Problem 1P , additional homework tip  4

Explanation

Calculation:

The graph of the function y=2.1cos0.52t is given by

Algebra and Trigonometry (MindTap Course List), Chapter 6.FOM, Problem 1P , additional homework tip  5

Final statement:

Algebra and Trigonometry (MindTap Course List), Chapter 6.FOM, Problem 1P , additional homework tip  6

To determine

(d)

To find:

The sine function that fits the data.

Answer

Solution:

y=2.05sin0.50t+1.55-0.01.

Explanation

Calculation:

Using the given data and the SinReg command on the T1-83 calculator, we get a function of the form y=asinbt+c+d.

Here, a=2.0487.2.05

b=0.5030790.50

c=1.551851.55

d=-0.008961-0.01

Apply the values in the function y=asinbt+c+d.

y=2.05sin0.50t+1.55-0.01.

Final statement:

y=2.05sin0.50t+1.55-0.01.

To determine

(e)

To find:

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

Answer

Solution:

y=2.05cos0.50t-0.02-0.01.

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

Explanation

Calculation:

From part (d), y=2.05sin0.50t+1.55-0.01.

Apply the reduction formula sinx=cosx-π2.

So, y=2.05cos0.50t+1.55-π2-0.01

y=2.05cos0.50t+1.55-1.57-0.01

y=2.05cos0.50t-0.02-0.01.

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

Final statement:

y=2.05cos0.50t-0.02-0.01.

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

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