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Find an Equation for the Sine or Cosine Wave - Problem 3
PhD. in Mathematics
Norm was 4th at the 2004 USA Weightlifting Nationals! He still trains and competes occasionally, despite his busy schedule.
I have a problem that’s related to finding the equation of a cosine or sine curve. Lisa records the depth of water in Springfield Harbor every two hours starting a midnight. Plot the data and find an equation that models it. I have the set of data here. This data is time in hours and depth in meters and I have a lot of data. I’ll plot as much of it as I need to, to identify the curve.
First of all the time goes up by increments of 2 hours. I need to figure out how I’m going to label my axis here. My horizontal axis will be time and I’ll make this point 4, 8, 12, 16, 20, 24. That will make it so it is possible to get all of my data on if I want to. The dependent variable goes from a minimum of 2 to a maximum of 10 rather. Let’s go 2, 4, 6, 8, 10 so there is 10, there is 2. Let’s find and plot all of the 10s, 2, and 14, that’s it. So at 2 we have 10 and at 14 we have 10. Now let’s plot all the 2's at 8 and at 20. Now let’s plot the 8's; 0, 4, 12, 16, and 24. And finally we’ve got the 4's; 6 and 10, 18 and 22. 6, 10, 18, 22. Let me plot this. This is my graph.
It’s not entirely clear whether this is a sign or cosine curve because it doesn’t start on its maximum the way a cosine curve would, nor does it start in the middle the way a sine curve would. So I’m not sure which this is. I actually could go either way.
What I’m going to do first is I’m going to say the easiest thing to do is to make it a cosine curve that shifted because I know where this point is. It’s exactly 2 units to the right of where a cosine curve would normally be. So I know that I can make this a shifted cosine curve, shifted 2 units to the right.
Let me start with the model y equals a times cosine of b times x minus h plus k. So this will have both kinds of shifts, horizontal and vertical and both kinds of stretches vertical, horizontal.
Fist of all what’s the A value? Well the maximum value is 10 and the minimum value is 2 so we know the amplitude is 10 minus 2 over 2, that’s 8 over 2. Which is 4. That make the A value either plus or minus 4. Now I’m going to guess that it’s plus 4 because I’m assuming that the graph’s been shifted 2 units to the right and the cosine curve normally starts on a maximum. Let’s make A equal 4. Let’s find the b value, that’s related to the period.
The period looks like, well it’s the distance between two consecutive maximums. We have two x equals 2 and x equals 14, that’s 12, the period is 12. We also know from the formula period equals 2 pi over b. So substituting 12 in, I get 12 equals 2 pi over b and I can solve for b. Multiply both sides by b. you get b times 12 equals 2pi. Then divide both sides by 12 and I get that b is pi over 6.
I already said that there’s a horizontal shift to the right, 2 units. That means h is 2. It remains to find what k is, the vertical shift. Well in order to find k,I need to find what the mid line of this graph is because normally the cosine curve is centered on the x axis. If we shift it up, that mid line also shifts up. So all I have to find is what the middle line is between the maximum and minimum values. K is the maximum plus the minimum divided by 2. The middle of those two values is the average. Now the maximum is 10 and the minimum is 2. So k is going is going to be 12 over 2 or 6. And now we’ve got k equals 6, h equals 2, b equals pi over 6 and a equals 4 and we can write our equation, y equals 4 times cosine of pi over 6 times the quantity x minus 2 plus 6.
That’s it and it’s got everything. It's got our vertical stretch, horizontal stretch, horizontal shift and vertical shift.
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