PRECALCULUS:...COMMON CORE ED.-W/ACCESS
PRECALCULUS:...COMMON CORE ED.-W/ACCESS
10th Edition
ISBN: 9780134781839
Author: Demana
Publisher: PEARSON
Expert Solution & Answer
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Chapter 5.1, Problem 85E
Solution

a)

To find: Draw the scatter plot of the given day and distance data.

The scatter plot will be as follows.

  PRECALCULUS:...COMMON CORE ED.-W/ACCESS, Chapter 5.1, Problem 85E , additional homework tip 1

Given information:

The given table is as follows.

    Date Day Distance (Mm)
    Mar 190403.9
    Mar 256383.0
    Mar 3112363.5
    Apr 618381.5
    Apr 1224400.9
    Apr 1830402.4
    Apr 2436371.0
    Apr 3042363.5
    May 648391.9
    May 1254405.4

Calculation:

Enter the day values on L1 and the distance values on L2 and make a scatter plot as shown below.

  PRECALCULUS:...COMMON CORE ED.-W/ACCESS, Chapter 5.1, Problem 85E , additional homework tip 2

Figure (1)

The scatter plot will be as follows.

  PRECALCULUS:...COMMON CORE ED.-W/ACCESS, Chapter 5.1, Problem 85E , additional homework tip 3

Figure (2)

b)

To find: The equation of best fit sine curve and superimposed it on the scatter plot.

The equation is y≈21.36sin0.227x+1.89+385 , and required graph is shown in figure (4).

Calculation:

Use the SinReg feature to find the sine regression model and enter it on Y1 and superimpose with the scatter plot.

The equation is,

  y≈21.36sin0.227x+1.89+385

The superimposed graph is as follows.

  PRECALCULUS:...COMMON CORE ED.-W/ACCESS, Chapter 5.1, Problem 85E , additional homework tip 4

Figure (3)

  PRECALCULUS:...COMMON CORE ED.-W/ACCESS, Chapter 5.1, Problem 85E , additional homework tip 5

Figure (4)

Therefore, the equation is y≈21.36sin0.227x+1.89+385 , and required graph is shown in figure (4).

c)

To find: The approximate number of days from apogee to apogee.

The approximate number of days from apogee to apogee is 27.3 days .

Calculation:

The number of days from apogee to apogee is the period of the sine regression which is 2πb . Since b≈0.227 from part (b), the period is:

  2πb=2π0.227≈27.7 days

Therefore, this is very close to the fact that the moon orbits the Earth once every 27.3 days .

d)

To find: The approximate distance at perigee.

The approximate distance at perigee is 363,500 km .

Calculation:

Use the minimum feature to find the perigee:

  PRECALCULUS:...COMMON CORE ED.-W/ACCESS, Chapter 5.1, Problem 85E , additional homework tip 6

The perigee is about 363.623 Mm or 363,623 km whereas it is 363.5 Mm or 363,500 km from the table.

Therefore, the approximate distance at perigee is 363,500 km .

e)

To find: The date on which on which Moon is 359000 km from Earth and also explain the reason.

The date is day 40 or April 28 .

Calculation:

The distance of 359,000 km happened somewhere between day 36 and day 42 which may be on day 40 or Apr 28 . The last five data points suggest a narrower sine model which can be used to find another sine regression supporting day 40 as the answer.

  PRECALCULUS:...COMMON CORE ED.-W/ACCESS, Chapter 5.1, Problem 85E , additional homework tip 7

  PRECALCULUS:...COMMON CORE ED.-W/ACCESS, Chapter 5.1, Problem 85E , additional homework tip 8

Therefore, the date is day 40 or April 28 .

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Chapter 5.1, Problem 84E
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Chapter 5.1, Problem 87E

Chapter 5 Solutions

PRECALCULUS:...COMMON CORE ED.-W/ACCESS

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