Finding Equations of Sinusoidal Functions From Real-World Data

Finding Equations of Sinusoidal Functions From

Real-World Data

**Note: Throughout this handout you will be asked to use your graphing calculator to verify certain results, but be aware that you will NOT be able to use your calculator during the exam!

Please make sure you have read through sections 2.1, 2.2, and 2.3 carefully because there will be references made to examples and terms defined in those sections. Also, this supplement will be using Example 4 and problems 11 and 13 from section 2.3 (pages 113 and116) from your text book, so make sure you have those in front of you.

In order to find the equation of a sinusoidal function from a set of given real-world data you must follow three steps.

Step 1: Make sure the data has a sinusoidal relationship by plotting the data in a graph.

Step 2: If the data shows a sinusoidal relationship, then begin finding the basic parameters of a sinusoidal function (i.e. – amplitude, period, etc.).

Step 3: Use the information from Step 2 to find an equation that fits the data. As a check on your work plot the equation on a graphing calculator to make sure it fits the data nicely (be aware that doing this by hand will NOT give you an equation that fits the data exactly, but it should appear to fit it nearly perfectly).

Steps 1 and 3 are easy, it is Step 2 that requires the most work. Our ultimate goal is to end up with an equation like ^{y Asin}_^{B x C}









we need to find A, B, C, and D.

Let us start by working through Example 4 on page 113. The data has already been plotted for you and it is fairly obvious that it represents a sinusoidal function. By replacing the months with numbers (January = 1 and December = 12)and using them as our x values, the table of given values is:

x 1 2 3 4 5 6 7 8 9 10 11 12

y 6.5 9 12 15 17.5 19 18 16 13.5 10 6.5 5.5

 First find the coefficient A. Remember the amplitude A of a sinusoidal function tells us the distance along the y-axis from the graph’s midline to its maximum or minimum, therefore ¹ Max. Min.

A  2  . For a pure sinusoidal wave such as sin

y x, this formula gives us ¹¹

 

2 2

A      , which is what we

value is 5.5, therefore ¹19 5.5 ¹13.5 6.75

2 2

A     . We can use either the

value 6.75 or the value 6.75 for our A. The final value we choose depends on the equation we decide to use at the end. Remember that a negative sign in front of a function reflects it across the x-axis (i.e. – the graph is turned “upside- down”).

 Next find B (in physics and engineering B is usually called the “radian frequency”). Remember that the period P is related by 2

P B

  , so in order to

find B we first need to find P (notice that B ² P

  ). There are several ways to do this. First we can measure the distance along the x-axis between two maxima (plural for maximum) or two minima (plural for minimum). You can visualize this by recalling the graph of ycosx, its first maximum occurs when x0and its next maximum occurs at x2. The distance between two points on a number line is x₂  x₁ , therefore the period is P 2 0 2 which agrees with the formula 2

P B

  since B1 for ycosx. What if we don’t have two

maxima or two minima such as with ysinx? After all, the data given above has only one maximum y19 at x6. If you refer back to section 2.1 (page 75), the first full paragraph emphasizes the fact that every cycle (or period) of a sinusoidal function can be divided into four arc sections of equal length. Now notice that between the maximum and minimum values there are two arc sections.

Figure 1: Two arc-sections (the solid portion) of y = sin x.

Since each arc section makes one-fourth of one period, two of them make one- half of one period. Therefore the distance from one maximum to the next minimum makes one-half of a period. Thus the period is twice this distance.

Let’s use ysinx as an example. The first maximum occurs at x 2

 and the next minimum occurs at ³

x_ 2

. The distance between these two points is

3 2

2 2 2

     . Multiplying the result by 2 gives us a period of 2 , which is what we expect.

Turning back to the problem at hand, our first maximum occurs at x6and our next minimum occurs at x12, and 12 6 6 so our period P is 12 which gives us a B of

6

 . Why we couldn’t use 12 from the start? After all, they gave us 12 points and the months repeat themselves every year. In the real-world you can’t assume that the data given to you will always represent one period. It may represent several periods, or even less than one period! Also, why not use 12? We can use a positive or negative value for B just like we can choose a positive or negative value for A, but using a negative value for B can sometimes complicate coming up with the appropriate equation. Thus, for the remainder of this handout and the remainder of the textbook, we will assume B0.

 Next let’s find the vertical shift D. D is the vertical midpoint between the maximum and minimum of a sinusoidal function. Remember that the midpoint between two points is their average ^{1 2}

2 y y

. For example, for either ysinxor cos

y x, the constant D is: ¹

 

2 0

D  

  , which is what we expect. For our problem, the maximum is 19 and the minimum is 5.5, therefore:

19 5.5 24.5

12.25

2 2

D 

   .

 Lastly, let’s find the phase shift (horizontal shift) C. This is the trickiest

parameter to find sometimes, and unlike A, B, and D there are an infinite number of choices possible for C! Of course we’ll only be focusing on finding a few of them. Again let’s refer back to page 75. We can interpret C as the “starting”

point (along the x-axis) for our graph. For ysinx this is the critical point one-

if we extend the graph backwards for negative values of x, it is also the point one- fourth of a period after the previous (or any) minimum (see Figure 3).

Figure 2: The critical point Figure 3: The critical point before a maximum. after a minimum.

Thus, for ysinx C0, but we could also use 2, 2 , 4 , and so on. To see for yourself, try graphing ysinx, ^ysin









 

sin 4

y x  on a graphing calculator to see that they all yield the same graph.

Note that this is only true if A0. If A is negative, then C is located one-fourth of a period before any minimum or one-fourth of a period after any maximum (this is simply the opposite of the case when A0). Graph y sinx to verify this. For ycosx the problem is much easier. Let the “starting” point of the graph of ycosx be its first maximum. Thus, we can find C by finding the first (or any) maximum and its horizontal distance from the origin





cos

y x C0 again, and we can also use the values 2, 2 , and 4 just like we did for ysinx (test this on a graphing calculator). Again, this is only true when A0. If A is negative, then C is the horizontal distance to any minimum. We’ll hold off on choosing a value for C for our problem for now.

The reason for this is that an appropriate value for C depends on whether we want to use a sine or cosine function and whether or not there are any restrictions such as (A0 and C0).

So let’s recap the information we’ve determined:

6.75 6

? 12.25 A

B C D











Now let’s go along with the book’s decision to use the form ^{y Asin}_^{B x C}





The book decides to use A = 6.75 so we’ll go along with that for now. Next we find C.

Remember that since we’re using the sine function and we’re using A > 0, so C must be located one-fourth of a period behind any maximum. The period of our function is 12, and one-fourth of 12 is 3. The maximum occurs at x6, which gives us:

Max. 1 6 3 3

C 4P   . Therefore our final answer is:

 

6.75sin 3 12.25 y _6 x _

    . This agrees with the book’s answer on page 114.

Alternatively we could have used the fact that C is one-fourth of a period after any minimum. The minimum occurs at x12, thus Min. ¹ 12 3 15

C 4P   . In this case our answer would have been: ^6.75sin





y 6 x  .

Let’s find one more alternative using a cosine function









6

C . Thus our answer would be: ^6.75cos





y 6 x  . Graph all three answers to see that they all yield the same graph.

Stop here and try working out problems 11 and 13 on page 116 on your own before moving on to the next page. Don’t be discouraged if you answers differ slightly in your choice of C from the answers given in the back of the book. Remember, there are an infinite number of choices for C. You can verify that your answer works by plotting your answer and the book’s answer in a graphing calculator. Once you’re done working out 11 and 13 see the next page for a detailed explanation of the book’s solution and alternative solutions

Solution(s) to Problem 11 from section 2.3 (page 116):

Problem 11

2, 21

5, 15

9.5, 10.5 11, 13

12, 16

15, 19 17, 21

22, 15

24.5, 10.5

8 10 12 14 16 18 20 22

0 3 6 9 12 15 18 21 24 27 30

x

y

We want to find an equation in the form ^yAsin_^{B x C}





list some important points:

There are two maxima at the points



 



There are two minima at the points



 



We can see that the data seems to follow a sinusoidal relationship.

See the next page for a summary of the information we can gather from the data.

From the given information we can find our parameters A, P (in order to find B and C), B, C, and D.

1 1

21 10.5 10.5 5.25

2 2

A    

So A = 5.25 (because the given restriction says A0)

17 2 15

P  

1 1

15 3.75

4P  4 (we need this value to find C)

2 2

B 15 P

 

 

Max. 1 2 3.75 1.75 C 4P   

21 10.5

15.75 D 2 

Therefore, the answer is: ^{5.25sin 2}





y 15 x 

Because the book gives no restriction on the phase shift C, we can use an alternative value by using the point one-fourth of a period after the first minimum.

Min. 1 9.5 3.75 13.25 C 4P  

Using this value, we come up with the alternative answer:

 

5.25sin 2 13.25 15.75 y 15 x  .

Now, just as some extra practice, let’s assume the restrictions were A0 and 0 C 24. In this case, P, B and D are the same, but A and C must change.

5.25 A 

5.25

A  (because of our restriction) 15

P and ¹ 3.75 4P 2

B_ 15



Max. 1 2 3.75 5.75

C 4P   (remember that we must reverse our formula for C when A0).

Thus, our answer is: ^{5.25sin 2}





y _15 x _

     . Another pair of equivalent answers are: ^{5.25sin 2}





y _15 x _

     and (using a cosine function)

 

5.25cos 2 9.5 15.75 y_{  }15 x_{ }

 

  . Do some extra practice by figuring out how to find these last two answers.

Solution(s) to Problem 13 from section 2.3 (page 116):

The data given in the book is:

x 1 2 3 4 5 6 7 8 9 10 11 12

y 73 71 74 79 82 85 87 90 87 83 79 74

Graphing this by hand should yield a graph similar to this one:

Problem 13

1, 73 2, 71 3, 74 4, 79 5, 82 6, 85 7, 87 8, 90 9, 87 10, 83 11, 79 12, 74

60 65 70 75 80 85 90 95 100

0 2 4 6 8 10 12 14

x

y

We want to find an equation in the form ^yAsin_^{B x C}





list some important points:

There is one maximum at the point





There is one minimum at the point





We can see that the data seems to follow a sinusoidal relationship.

From the given information we can find our parameters A, P (in order to find B and C), B, C, and D.

1 1

90 71 19 9.5

2 2

A    

9.5

A (because of the restriction on A) 2 8 2 2 6 12

P    (our max. and min. occur at x = 8 and x = 2)

1 1

12 3 4P 4 

2 2

12 6 B P

  

  

Max. 1 8 3 5 C 4P  

90 71 161 2 2 80.5

D   

Therefore, the answer is: ^9.5sin





y 6 x  .

Extra practice: Do problem 13 again but this time assume the restriction is A < 0. See if you can come up with these answers:

Answer 1: ^9.5sin





y  6 x  Answer 2: ^9.75sin





y  6 x 

Extra practice: Do problem 13 again but this time use a cosine function and find one answer where A > 0 and another where A < 0 and see if you can come up with these answers.

 

 

Answer 2: ^9.5cos





y  6 x 

**Optional: Here’s a neat trick to quickly find another phase shift C for any given problem (that is after you have already found one value). Now think about the graph of sine or cosine, remember that we considered C to be our “starting” point (on the x-axis) for the graph. Notice that if we extend the graph in either direction, these points repeat themselves every cycle/period of the graph.

Thus we can use the formula: C_new C_old k P where C_old is the first value of C you found the traditional way, Cnew is your new value for C, k is any integer (remember the integers are the numbers ... 2, 1, 0, 1, 2...) and P is the period of your function. Try this out with any of the problems worked out in this handout find a family of equivalent functions with different values of C. Remember in order to check if they’re equivalent you can use your graphing calculator to graph them. If all the functions have the same graph, then they are equivalent.

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