GTF - Modeling with Trigonometric Functions Lesson

Modeling with Trig Functions

In this lesson, we will explore how to write a sinusoidal function given a graph, a description or even a real-world situation.

 

So, let's look at what we know about this graph:

Recall that a sine function could potentially have 4 transformations, represented by A, B, C and D in the equation: $f\left(x\right)=A\sin\left(Bx+C\right)+D$

1. The maximum and minimum are equidistant from the x-axis, which means that the midline is y = 0, so D = 0.
2. The distance from the midline to the maximum or minimum is 2, so A = 2
3. The period is $2\pi,\:$because that is how long it takes for the curve to repeat itself. Since the period is $2\pi,$ we know that B = 1.
4. We know that sine starts at the midline and goes up, so we can identify the starting points as $\left(\frac{\pi}{2},\:0\right)$, so, then we can solve for C: $-\frac{C}{B}=\frac{\pi}{2} \\ -\frac{C}{1}=\frac{\pi}{2} \\ {C}=-\frac{\pi}{2} \\$
5. So, now that we have found A, B, C and D - we can write our equation: $f\left(x\right)=2\sin\left(x-\frac{\pi}{2}\right)$

 

Watch this video to practice a few more.

 

 

 

 

Let's think about some real-world scenarios.

Often, average daily high temperatures will follow a sinusoidal pattern (recall: both sine and cosine curves are considered sinusoidal!). In Plainville, GA, the average daily temperature in January was a minimum of 40 degrees. The city hit the highest average daily temperature in July at 90 degrees. Write a sine or cosine function to represent this situation.

 

1. Let's start by plotting the information we know. I am going to let the x-axis represent the time in months since January, and let January be t=0. So, the point (6, 90) represents the high temperature in July.

 

 

2. I have not graphed a full period yet, since the curve starts at a minimum. In order to complete the period, I need to go ahead and plot the next minimum point at (12, 40). Since the x-axis represents months since January – this point represents hitting the low temperature again in January.

 

3. Now we can see that the period is 12 months, so we can solve for B.

$\frac{2\pi}{B}=12 \\ {2\pi}=12B \\ \frac{\pi}{6}=B$

4. Let's determine the amplitude and equation of the midline next. The maximum and minimum values are 90 - 40 = 50 units apart, so the
amplitude = 50/2 = 25. This means this midline is y = 40 +25 = 65 or  y = 90 - 25 = 65.  So we have:

|A| = 25

D = 65

 

5. Now, we should decide if this represents a sine or cosine curve, this will help us decide if there should be a phase shift. Since this curve starts at a minimum (and not at the midline), we should let it be a cosine curve. Cosine starts at a maximum, so if we multiply A by -1, we will reflect it over the midline and start at the minimum to match our situation.

6. Now, we've found our equation and we can graph it to be sure it fits our points: $D\left(t\right)=-25\cos\left(\frac{\pi}{6}t\right)+65$

 

Try this problem:

 IMAGES CREATED BY GAVS