Big Ideas Math Algebra 2, 2014
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Big Ideas Math Algebra 2, 2014 View details
6. Modeling with Trigonometric Functions
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Exercise 31 Page 512

Practice makes perfect
a We are told that the low tide at a port is 3.5 feet and occurs at midnight, which is represented by t=0. After 6 hours the port is at high tide, which is 16.5 feet.

We want to write a sinusoidal model for the tide depth d at the port, as a function of the time t. To do so, we will consider the general forms of transformed sine and cosine functions. Sine Function:& y= asin b(t- h)+ k Cosine Function:& y= acos b(t- h)+ k In these functions | a| is the amplitude, 2Ď€ b is the period ( b>0), h is the horizontal shift, and k is the vertical shift. To write a function for a sinusoid, we will follow four steps.

  1. Use the maximum and minimum values to identify any vertical shift k.
  2. Determine whether the tide depth should be modeled by a sine or a cosine function, and identify any horizontal shift h.
  3. Find the period and the value of b.
  4. Find the amplitude and the value of a.
Let's do it!

Step 1

The vertical shift k is the mean of the maximum and minimum values. k=maximum value+ minimum value/2 We know that the maximum value is 16.5 and the minimum value is 3.5. k=16.5+ 3.5/2 Let's evaluate this formula to find the value of k.
k = 16.5+3.5/2
k = 20/2
k= 10
We found that the vertical shift k is 10.

Step 2

Since the minimum point occurs at t=0, the function passes through y-axis at its minimum point. Then, the tide depth can be modeled as a cosine function reflected across the x-axis and with no horizontal shift. This means that h= 0.

Step 3

We know that six hours after midnight the port is at high tide. This represents the half of one cycle. To find the period we need to multiply six hours by two, because period corresponds to one whole cycle. Period: 2(6) = 12 We also know that the period of sine and cosine functions is 2Ď€ b. With this information, we can write an equation in terms of b. 12=2Ď€/b Let's solve this equation!
12 = 2Ď€/b
â–Ľ
Solve for b
12b =2 π
b = 2Ď€/12
b = π/6
We found that b= π6.

Step 4

Next, let's find the amplitude | a|. This is half the difference between the maximum and minimum values. | a|=maximum value- minimum value/2 In our case, the maximum value is 16.5 and minimum value is 3.5. | a|=16.5- 3.5/2 We will solve this equation to find | a|. Let's do it!
|a| = 16.5-3.5/2
|a| = 13/2
|a| = 6.5
Since the graph is a reflection in the x-axis of cosine function, we know that a<0. Therefore, a= - 6.5.

Equation of the Model

Because of the location of the minimum point, we concluded that the tide depth should be modeled by a cosine function. We also found that a= - 6.5, b= π6, h= 0, and k= 10. Finally, we can write the equation of the model. d = - 6.5 cos π/6( t - 0)+ 10 ⇕ d=- 6.5 cos π/6 t + 10

b Recall the known formulas for the x -intercepts, maximum values, and minimum values of a cosine function of the form y= a cos bx when a>0. In our case a= - 6.5 and b= π6, so the graph is reflected in the x -axis. Therefore, the minimum becomes the maximum and the maximum becomes the minimum.
Formula
x-intercepts (1/4*2Ď€/b,0), (3/4*2Ď€/b,0)
Minimum (0, a), (2Ď€/b, a )
Maximum (1/2*2Ď€/b, - a)
Our function is in the form y= acos b(x- h)+ k. The value of k tells us that our graph is a vertical translation. In this case it is a translation 10 units up. Thus, we add 10 to the y -coordinates of the points from the table. Since h= 0, there is no horizontal translation, so we do not change the x-coordinates.
x-intercepts Minimum Maximum
(1/4*2Ď€/b, k), (3/4*2Ď€/b, k) (0, a+ k), (2Ď€/b, a + k) (1/2*2Ď€/b, - a+ k)
(1/4*2Ď€/Ď€6, 10), (3/4*2Ď€/Ď€6, 10) (0, - 6.5+ 10), (2Ď€/Ď€6, - 6.5+ 10 ) (1/2*2Ď€/Ď€6, - ( - 6.5)+ 10)
(3,10), (9,10) (0,3.5), (12,3.5) (6,16.5)

Now we know five points that the given function passes through. Let's graph the function by plotting and connecting them with a smooth curve. Let's do it!

On the graph we can identify the maximum and minimum values of the function in a 24-hour period. These values will represent the high and the low tide at the port.

As we can see, the minimum is at 12 and 24 and the maximum is at 6 and 18. This means that the low tide is at 12:00AM and 12:00PM and the high tide is at 6:00AM and 6:00PM.

c Since our model from Part A represents the tide depth after midnight t=0, to find a graph that shows the tide depth at the port t hours after 3:00AM we need to add 3 to the argument t. Graphically this will represent a horizontal shift of the parent function obtained in Part A by 3 units to the left.

d=- 6.5 cos π/6t+ 10 ⇓ d=- 6.5 cos π/6(t + 3)+ 10 Let's show this on a coordinate plane!