There are many situations in the work-a-day world where a model that uses trigonometry is appropriate. Tides, time of sunrise, wave motion, vibrations, and many other phenomenon can be modeled using trigonometric functions. Let's use trigonometry to model a couple of situations.
Example 1 - Suppose we have a body that moves between 3 inches above a midline and 3 inches below the midline. Let us further suppose that starting on the midline at time 0 it takes our body 2 seconds to reach a peak, another 2 to be back on the midline, 2 more seconds to reach a low point and at last 2 more to return to the midline. The process goes on forever. Can we model this with trigonometric functions? Yes we can!
- First we note that the time for a complete cycle is 8 seconds. That is the period of our process. Using the fact that the period (P) is equal to 2*Pi over B we can solve for B to get B = [2*Pi]/8 = Pi/4.
- Next let's decide if we have a sine or cosine function. At time t = 0 the body is at the centerline (origin). If the function is not translated horizontally the sine function starts at the midline while the cosine function starts off the midline. It would appear that ours is a sine function.
- The body goes 3 inches above the midline and 3 inches below the midline. The amplitude of our function must be 3.
- Based on what we know our function is: y = 3*Sin(Pi/4 * t). The figure on the right shows one cycle of motion. This motion could describe a 3 inch long shaft in your car that starts horizontally and rotates in a counterclockwise dierction.
Example 2 - A power company serves two different cities. Let P1 be the power requirements of city 1 and P2 be the power requirements of city 2. Both P1 and P2 are calculated in terms of hours elapsed since midnight. The power requirements are defined as follows:
Power Requirements of city 1 and 2
If we look at the two power requirements we can easily tell that the peak power requirement for city 1 is 120 megawatts and if occurs at 600 :AM. The peak power requirement for city 2 is 130 megawatts and it occurs at midnight. If we want to determine the total load on the power company's system we need to combine the two functions:
- P1 = 80 - 40 sin(p*t/12)
- P2 = 100 + 30 cos(p*t/12)
- P1 + P2 = 80 - 40 sin(p*t/12) + 100 + 30 cos(p/12 * t)
- P1 + P2 = 180 - (40 sin(p*t/12 ) - 30 cos(p*t/12))
- A = [ (40)2 + (-30)2]1/2 = 50
- Tanf= (-30/40) = -0.75, Fourth Quadrant.
- f = 0.6435
- P1 + P2 = 180 - 50 sin(p*t/12 - 0.6435).
- The peak power requirement will be 230 megawatts and it will occur at approximately 8:30 PM (hour 20.5).