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Find an Equation for the Sine or Cosine Wave - Problem 2
PhD. in Mathematics
Norm was 4th at the 2004 USA Weightlifting Nationals! He still trains and competes occasionally, despite his busy schedule.
I’m finding the equation of the graph of a sine or cosine function and I have one here. This function, I first need to figure to whether I want to call it a sine or a cosine curve and I think I want to call it a cosine because it starts at a maximum here. But here’s a little problem, this thing is not centered on the x axis the way I usually expect sine and cosine curves to be, so it looks like it's been shifted up. And that means that if it’s a cosine curve that’s been shifted up, I’m going to have an equation of the form y equals A cosine bx plus k. That plus k is going to give me that upward shift that I need.
First of all let’s figure out what the amplitude is. Remember that amplitude is the maximum minus the minimum over 2. This is the way amplitude is defined. The maximum value based on the points that I’m given is 15 and the minimum value is negative 5. So 15 minus -5 over 2 and that’s 20 over 2 which is 10. My A value is going to be either +10 or -10 depending on whether or not I think there is a reflection, but I don’t think there is. This is the maximum value, that’s where cosine usually starts, so I’m going to say that it’s a positive 10. A is 10.
Now I should find out what b is. Now b depends on the period of this function. Judging by my graph, the period is 12pi. It’s the time it takes to go from one maximum to the next, 12pi. So if the period is 12 pi, then we use the formula for period, 2pi over b and I’ll substitute 12pi for the period. 12pi equals 2 pi over b. So I solve this for b.
First multiply both sides by b and then divide both sides by 12pi and I get b equals 1/6. I’ve got A and b, how do I get k, the vertical shift?
Like I said the sine and cosine graphs are normally centered on the x axis, that’s their midline. It’s the line that’s right halfway between the maximum value and the minimum values, whereas now the midline looks like it's right here. This midline is exactly halfway between he maximum and minimum values. How do we find a number that’s exactly halfway between 2 numbers? We average them. So this midline y equals k is going to have a k value that’s exactly equal to the average of the maximum and the minimum, so 15 plus -5 over 2. That’s 10 over 2 which is 5. My k value is 5.
My final equation is y equals 10 cosine 1/6x plus 5. Remember if you have a vertical shift use the formula, k equals max plus min over 2 and you’ll find what you’re vertical shift is.
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