III
III. Solutions to the lesson:
Applying Trigonometric Functions to Everyday Life Situations
Regressions
1. Based on the shape of the graph, what type of regression will be appropriate? (Hint: You might want to anticipate what the data might look like for the next year or two.) Explain your decision by providing 2 aspects of the graph that are unique to this type of equation.
Based on the data, it appears that a Sinusoidal regression would fit best. The follows an “s”-shape or snake shape. It must be a sinusoidal equation because the data will repeat year after year. That is why it must be a sinusoidal equation.
[pic]
2. Write your regression equation. How well does it fit the data? Explain.
y = 183.379 sin( 0.016749x – 1.32212) + 727.711
By simply evaluating the regression by examining the graph, it appears that the equation fits the data well.
3. Find the Domain and Range of the sine curve.
Domain: All real numbers
Range: [544.32, 911.09]
In this activity, we will find the period, amplitude and the phase shift of a sinusoidal curve. Before we can accomplish this task we must expand our view of the sinusoidal regression because we can only see the curve for one year. Let’s expand the viewing window by raising the x-max to 3 years or 1100 days. Now, we can see that the curve repeats for each calendar year.
Attributes of a Sinusoidal Graph Questions:
Period:
Now, let’s discuss the period of a curve. The period is amount of independent values that are required for the dependent values repeat or to complete one cycle. To find the period of our sine regression, we need to find two days that have the same length in time.
4. The easiest values to locate on a periodic graph are the maximums and minimums. Use your technology to find the first two maximums or the first two minimums.
Maximums: (172.719, 911.09), (547.85, 911.09) or
Minimums: (360.284, 544.332), (735.415, 544.332)
[pic]
5. Subtract the two consecutive x-values to find the period. What is the period? How do you know that your answer is correct?
Maxs: 547.85 – 172.719 = 375.131
Mins: 735.415 – 369.284 = 375.131
The answer should be 365.25 days because that is the period of the calendar year. However, the period found is very close to what it should be. It is just a regression. It may not match the data exactly.
6. If you found the maximum values, find the minimum values or vice versa.
See above.
7. Write the days that give you the shortest and longest days. Using the Old Farmer’s Almanac tables, change them from days of the year to days of each month. What do you notice about each day? Why are these days familiar?
Longest (911.09mins): 172.719 ( June 21st
547.85 – 366 = 182 ( June 30th
922.981 – (366 + 365) = 191.981 ( July 9th
June 22nd, the summer solstice, or the first day of summer.
Shortest (544.332): 360.284 ( December 25th
735.415 – 366 = 369.415 ( January 4th
The days are close to December 21st, which is the winter solstice, or the first day of winter.
8. Why does the regression equation not give us the exact dates?
The regression equation is just a mathematical model that follows that data. It is not the physical universe so it does not generate the data. It appears that as the number of years exceeds the year 2008, the values from the regression equation become less accurate.
9. What is the period of the function [pic]
The period for f(x) = sin(x) is 2π.
10. The period of the regression equation is different than [pic]. Right now, it cannot be seen in the equation [pic], but it’s there. To find it we must find the ratio of the period of sin(x) and the period of the regression equation. Where is this ratio in the regression equation?
[pic]This value of the b-coefficient.
11. So, describe how you could work backwards using the equation [pic] to find the period of any sine equation?
To find the period of a sine curve using the equation, divide 2π by the b-coefficient.
Amplitude:
Next, we want to investigate the amplitude of a sinusoidal curve. Generally, amplitude is defined as height of the curve above a horizontal axis. Usually the x-axis is the horizontal axis. Here, the regression curve certainly exceeds y = 0. So, we will find the amplitude by saying that it is half of the difference between the maximum and minimum of the graph.
12. Using the values of the local maximums and minimums that you found above, write equations of the horizontal lines that bound the curve.
The equations of the horizontal lines that bound the curve are y = 544.332 and y = 911.09.
13. Use these two lines to find the height of the curve and divide by 2. What is the amplitude of the graph?
[pic]is the amplitude of the graph.
14. Now, look at the regression equation [pic]. Do you see the amplitude in the equation? What coefficient a, b, c, or d is it?
The amplitude of the graph is the a-coefficient.
Phase Shift:
Finally, let’s consider the phase shift. The phase shift is the horizontal shift left or right. So, we need to recall where the function f(x) = sin(x) crosses the y-axis.
15. Using your knowledge of the graph of [pic], does it cross the y-axis nearer a minimum, a maximum, or halfway between each? Explain?
The function [pic]intersects the y-axis half way between a minimum and a maximum. It could be found by finding the midpoint of these two ordered-pairs.
16. We need to find this same point on our regression. To do this, change your window to see the minimum value just left of the y-axis. Find the point that was discussed in the previous question.
[pic] This value could also be found using geometry technology.
17. How far right of the y-axis is this point? How would you describe the phase shift?
The phase shift is 78.936 days to the right.
18. Do you see this value in the equation?
No.
19. This value is not in the equation because it is affected by the period/horizontal stretch. It would usually be the value subtracted from x. However, the coefficient of x changes this value. So, we need to factor [pic]into [pic]. What is the result? What do you notice about the value subtracted from x?
[pic] The value subtracted from x is the phase shift to the right.
20. What specific day of the year does this phase shift represent? What is the significance of this date?
78.9372 days of the year represents March 18th, which is the spring equinox. Here, we find the day of the year which has the mean/median length of day.
Vertical Shift:
Finally, let’s consider the vertical shift. The vertical shift is the shift up or down. So, we need to recall where the function f(x) = sin(x) crosses the y-axis.
21. Using your knowledge of the graph of [pic], does it cross the x-axis nearer a minimum, a maximum, or halfway between each? Explain?
The function [pic]intersects the y-axis half way between a minimum and a maximum. It could be found by finding the midpoint of these two ordered-pairs.
22. We need to find this same point on our regression. To do this, change your window to see the minimum value just left of the y-axis. Find the point that was discussed in the previous question. (This is the same answer to #2 to find the Phase Shift.)
[pic] This value could also be found using geometry technology.
23. How far above of the x-axis is this point? How would you describe the vertical shift?
727.711 minutes. This is approximately the length of March 18th , which is the spring equinox. Here, we find the day of the year which has the mean/median length of day.
24. Where in the equation [pic]do you see this value? a, b, c, or d?
727.711 is the d-coefficient.
Generalize your results:
Using a sine equation, [pic], explain how to find the following:
25. Period – divide 2π by the b-coefficient
26. Amplitude – it is the a-coefficient
27. Phase Shift – divide the opposite of the c-coefficient by the b-coefficient
28. Vertical Shift- a) The horizontal distance a periodic function is shifted from its normal x-intercept
b) It’s the d-coefficient
Numerical Analysis:
Tangent Lines and Rates of Change.
In this portion of the lab, you will explore how rates of change, as observed in a table, can be used to find interesting points on the graph of a function.
29. As an appendix to this activity, there are several charts featuring the day of the month, the day of the year, and the length of daylight in minutes. Please refer to the table showing the entire year. What general trend do you observe in the amount of daylight from the beginning of the year to the end?
The daylight increases into the month of June, and then decreases through December.
30. Make an estimate on about what month and day of the year that each of the following occurs. [Hint: The answers for the questions below may not occur on a specific table value date, but rather in-between values]:
a. Maximum amount of daylight.
About June 20 or thereabouts.
b. Minimum amount of daylight.
About December 20 or thereabouts.
c. Greatest increase in the amount of daylight.
About March 20 or thereabouts.
a. Greatest decrease in the amount of daylight.
About September 20 or thereabouts.
31. Now examine the table featuring the values for the month of June.
d. Using information from day 1 and day 3 of June, find the rate of change in the daylight per day.
i. Subtract the daylight for day 1 from day 3.
908 – 905 = 3 minutes
ii. Divide this value by 2 days (since the difference in the days is 2).
3/2 = 1.5
iii. What are the units of this quotient? minutes per day
e. Repeat steps i) and ii) for the following days: Days 18 and 20.
917 – 917 = 0
0/2 = 0 minute per day
f. Repeat the steps from above by subtracting the daylight from day 28 from day 30.
914 – 915 = -1 minute per day
g. Explain what it means for answer for a) to be positive, the answer for b) to be zero, and the answer for c) to be negative.
a) is positive because the amount of daylight in early June is increasing.
b) is zero because the amount of daylight is neither increasing nor decreasing.
c) is negative because the amount of daylight in late June is decreasing.
h. What you found in part b) above is known as the “summer solstice,” and marks the “longest day of the year,” or the day on which the amount of daylight reaches a maximum (the “winter solstice,” or “shortest day of the year,” happens exactly six months later in December.) For 2008, the summer solstice falls on June 21. How close was this to your guess from part 2 a) from the previous page?
See student answers.
i. What else changes or “begins” on the summer solstice? On the winter solstice?
Summer begins on the summer solstice. Winter begins on the winter solstice.
32. Now examine the table featuring the values for the month of March.
j. Using information from day 1 and day 3 of March, find the rate of change in the daylight per day.
i. Subtract the daylight for day 1 from day 3.
681 – 676 = 5 minutes
ii. Divide this value by 2 days (since the difference in the days is 2).
Be sure to include proper units.
5/2 = 2.5 minutes per day
k. Repeat steps i) and ii) for the following days: Days 19 and 21.
733 – 727 = 6 minutes
6/2 = 3 minutes per day
l. Repeat the steps for the following days: Days 29 and 31.
762 – 757 = 5 minutes
5/2 = 2.5 minutes per day
m. Which of the above has the highest value?
Part h) has the highest value at 3 minutes per day.
n. What you found in b) above is known as the “spring equinox,” and marks the day on which the rate of change of daylight reaches a positive maximum (the “autumnal equinox,” or “shortest day of the year,” happens exactly six months later in September.) For 2008, the spring equinox falls on March 20. How close was this to your guess from part 2 a) from two pages previous?
See student answers
o. What else changes or “begins” on the spring equinox? On the autumnal equinox?
Spring begins on the spring equinox. Autumn begins on the autumnal equinox.
33. One major concept that you will see in a future calculus course is that of a tangent line to a curve. The “tangent line” problem goes back to the time of Sir Isaac Newton, when he and other mathematicians were trying to understand how a tangent line function could be defined at any point on a curve.
p. Consult the table for the month of June. Using the data from June 19, create an ordered pair featuring (day of the year, amount of daylight)
(171, 917)
q. Using [pic], where m is the value of the rate of change of daylight on June 19, (which you calculated in 3 b)), create a tangent line equation.
Equation: [pic]
r. What kind of line is this tangent line (how would you describe its orientation)?
Horizontal; straight across; accept similar
s. On what other day of the year would this kind of line appear on a graph of amount of daylight vs. day of the year? Why?
Should be six months later on the winter solstice, or about December 20. This is because the winter solstice marks the “shortest day of the year,” when the year is neither gaining nor losing daylight.
t. Now consult the table for the month of March. Using the data from March 20, repeat steps a) and b) from above. The m will be the rate of increase of daylight for this day, which you found in 4 h) above.
(80, 730)
Equation:[pic]
u. On what other day of the year would this kind of line, featuring the same slope, except negative, appear on a graph of amount of daylight vs. day of the year? Why?
Should be six months later on the autumnal equinox, or about September 20. This is because the equinoxes are the days on which the rate of increase (or decrease, as in the case of the autumnal equinox) is at a maximum.
While pushing your nephew on a swing in the local park, you study how the swing goes back and forth and begin to think of your work with trigonometry in math class. Because you know that the chains that hold the swing are fastened at the center of a circle so that the motion of the swing is circular, you surmise that the distance of the swing from the center point of its motion is actually a cosine function. You decide that your nephew is not going to tire of the swing any time soon, so you take out your trusty tape measure and stopwatch and record the following times and distances:
|Time in seconds |0 |1 |2 |3 |4 |5 |
|Distance in feet |4 |0 |-4 |0 |4 |0 |
34. On graph paper, plot the points putting time on the x-axis and distance on the y-axis.
35. Since the swing takes four seconds to complete a cycle, the period of our function is four. Calculate the B value using the fact that 2(/B = Period
B = (/4
36. Calculate the amplitude of the function by finding the distance from the smallest and largest distance and dividing by 2.
Amplitude = [pic]( A ( = 4
37. Since the highest distance occurs when time = 0, is there any need to consider a phase shift of this cosine function?
No!
38. Write the equation of the function.
y = 4cos((/4)x
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