Algebra II (2018 edition)
Sal introduces the main features of sinusoidal functions: midline, amplitude, & period. He shows how these can be found from a sinusoidal function's graph. Created by Sal Khan.
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- Hello! How do I determine if a function has a period algebraically? Thanks!(42 votes)
- There is a way to do this, but to be honest it is much easier to do graphically. Also, the math involved can get fairly advanced and rather hard to avoid making errors with.
But here is how you would do it:
The function f(x) is periodic if and only if:
f(x+nL) - f(x) = 0, where n is any integer and L is some constant other than 0.
If the only solution for L is 0, then the function is NOT periodic.
Thus, set n=1 and solve for L.
After doing so, demonstrate that
f(x+nL) - f(x) = 0, for integer values of n.
So, that is how you would determine this mathematically. I don't recommend attempting it because it is quite difficult and often involves nonreal complex exponents or complex logarithms.
Note: there are some functions that have more than one period, but these are really advanced level math and you probably won't encounter them at this level of study.(62 votes)
- How do I know whether I must use midline = (max val + min val) / 2 or (max val - min val) / 2? I'm really confused(13 votes)
- Instead of relying on formulas that are so alike that they're confusing (to me, too!), maybe try to think it through each time (at least in the beginning) until it gets more familiar). I had a LOT of difficulty with this type of problem and I found that I had to go slowly and think things through each step EVERY time I did a problem.
Here's a method I found helpful. Maybe it will be of use to you.
I know that the midline lies halfway between the max and the min. So I need to get the total height (by subtracting the min from the max). (That gives me ( 4 - (-2) ). Now I divide by 2. (That gives me 3.) By definition that is the AMPLITUDE. Now I can either add that to the min (or subtract it from the max), and where I end up is the MIDLINE ( at 1 ).
Good luck!(26 votes)
- Can someone please explain how to find the midline of a sinusoidal function from its equation, instead of the graph?(9 votes)
- y = A sin (B(x - C)) + D is a general format for a sinusoidal function. The number in the D spot represents the midline. The equation of the midline is always 'y = D'.
y = 3 sin(2(x - π)) - 5 has a midline at y = -5(18 votes)
- What are sinusoidal functions?(6 votes)
- A sinusoidal function is one with a smooth, repetitive oscillation. "Sinusoidal" comes from "sine", because the sine function is a smooth, repetitive oscillation. Examples of everyday things which can be represented by sinusoidal functions are a swinging pendulum, a bouncing spring, or a vibrating guitar string.
I hope this helps. Good luck!(22 votes)
- Hello, I'm just wondering why Sal choice to use the Midline to find the period: is this always the case? or is it just easier to use the Midlines y value instead?
Thank you!(5 votes)
- Hi Daniel,
No, you do not have to use the midline to find the period. You can find the period by going from peak to peak, or trough to trough, or midline to midline. If you use midline of course you will need to keep in mind that you will need to skip a midline (because the midlines you measure from must be going the same direction).
Hope this helps,
- Convenient Colleague(14 votes)
- Do you have any videos that actually talk about the graphs of trig functions? If so please post as soon as possible. This title is very misleading. I assumed you would teach what the trig functions looked like but it seemed more like you expected us to know it already :(. On the next video I was so frustrated because I did not even know what -1/2 cos(3x) meant. I didn't even know these things could be graphed. I thought you only used for triangles or something. What is all this graphing stuff? SO frustrated :/(6 votes)
- If you watch the videos in the preceding section headed "Unit circle definition of trig functions", you will appreciate that the cosine and sine functions take an angle as the input value, and give output values that repeat every so often, and that always remain within the values -1 and 1. If, instead of thinking about the x and y coordinates of points on the unit circle, you decide to plot a graph with angle on the x-axis, with the y axis being the cosine or sine of the variable x, you will obtain a pattern like the one in this video.
So, this is the video where Sal is showing you what the trig functions look like. If the maximum value of the cosine or sine of any angle is 1, and the minimum value is -1, then the amplitude of these functions is 1, and any function that is a multiple of one of these functions will have an amplitude of 1 times that multiple, or -1/2 in the case of cos(3x).
Edit: Actually, all this is made more explicit in this video: https://www.khanacademy.org/math/trigonometry/trig-function-graphs/trig_graphs_tutorial/v/we-graph-domain-and-range-of-sine-function(5 votes)
- Can the "midline" also be called the "sinusoidal axis"?(4 votes)
- How is the wavelength of a wave(distance from crest to crest or trough to trough) related to the period? Is the wavelength of a sinusoidal function equivalent to the period? Sorry for including a bit of science here, I'm just confused.(3 votes)
- They aren't the same. Wavelength refers to the distance between two troughs/crests while period refers to the time taken to complete one wavelength(5 votes)
- I have watched this video over and over and i get amplitude and midline but finding the period makes no sense to me. is there a formula i can use?(3 votes)
- Periods of a sinusoidal functions are very very confusing so I can empathize with you on that.
Let's just say the given is from the midline to maximum, with a distance of 3.
Whenever you are given a mid-line to a maximum/minimum, always multiply that distance by 4. A graphic in the practice problems explains why.
3*4 = 12. Now for every time you want to find the period, use this formula.
2pi / (that number you multipled by 4).
//always use this formula when finding the period !
So now you have 2pi/12. Simplifying that, you get pi/6. boom, period !
Also if you have given like a maxiumum to maximum or minimum to minimum, instead of multiplying by 4, multiply by 2. Again the graphic shows a visual interpretation.
Hopefully that helps ! This is how I interpreted it as.(3 votes)
- For the Period of sinusoidal functions from graph activity, I graph the same extremum and midline point but my waves look different, therefore I get the question wrong, do you know how to fix this issue?(2 votes)
- Make sure that you are in the right mode. Use degree mode if the question asks for degrees and use radians if the questions asks for radians.(4 votes)
We have a periodic function depicted here and what I want you to do is think about what the midline of this function is. The midline is a line, a horizontal line, where half of the function is above it, and half of the function is below it. And then I want you to think about the amplitude. How far does this function vary from that midline-- either how far above does it go or how far does it go below it? It should be the same amount because the midline should be between the highest and the lowest points. And then finally, think about what the period of this function is. How much do you have to have a change in x to get to the same point in the cycle of this periodic function? So I encourage you to pause the video now and think about those questions. So let's tackle the midline first. So one way to think about is, well, how high does this function go? Well, the highest y-value for this function we see is 4. It keeps hitting 4 on a fairly regular basis. And we'll talk about how regular that is when we talk about the period. And what's the lowest value that this function gets to? Well, it gets to y equals negative 2. So what's halfway between 4 and negative 2? Well, you could eyeball it, or you could count, or you could, literally, just take the average between 4 and negative 2. So 4-- so the midline is going to be the horizontal line-- y is equal to 4 plus negative 2 over 2. Just literally the mean, the arithmetic mean, between 4 and negative 2. The average of 4 and negative 2, which is just going to be equal to one. So the line y equals 1 is the midline. So that's the midline right over here. And you see that it's kind of cutting the function where you have half of the function is above it, and half of the function is below it. So that's the midline. Now, let's think about the amplitude. Well, the amplitude is how much this function varies from the midline-- either above the midline or below the midline. And the midline is in the middle, so it's going to be the same amount whether you go above or below. One way to say it is, well, at this maximum point, right over here, how far above the midline is this? Well, to get from 1 to 4 you have to go-- you're 3 above the midline. Another way of thinking about this maximum point is y equals 4 minus y equals 1. Well, your y can go as much as 3 above the midline. Or you could say your y-value could be as much as 3 below the midline. That's this point right over here, 1 minus 3 is negative 1. So your amplitude right over here is equal to 3. You could vary as much as 3, either above the midline or below the midline. Finally, the period. And when I think about the period I try to look for a relatively convenient spot on the curve. And I'm calling this a convenient spot because it's a nice-- when x is at negative 2, y is it one-- it's at a nice integer value. And so what I want to do is keep traveling along this curve until I get to the same y-value but not just the same y-value but I get the same y-value that I'm also traveling in the same direction. So for example, let's travel along this curve. So essentially our x is increasing. Our x keeps increasing. Now you might say, hey, have I completed a cycle here because, once again, y is equal to 1? You haven't completed a cycle here because notice over here where our y is increasing as x increases. Well here our y is decreasing as x increases. Our slope is positive here. Our slope is negative here. So this isn't the same point on the cycle. We need to get to the point where y once again equals 1. Or we could say, especially in this case, we're at the midline again, but our slope is increasing. So let's just keep going. So that gets us to right over there. So notice, now we have completed one cycle. So the change in x needed to complete one cycle. That is your period. So to go from negative 2 to 0, your period is 2. So your period here is 2. And you could do it again. So we're at that point. Let's see, we want to get back to a point where we're at the midline-- and I just happen to start right over here at the midline. I could have started really at any point. You want to get to the same point but also where the slope is the same. We're at the same point in the cycle once again. So I could go-- so if I travel 1 I'm at the midline again but I'm now going down. So I have to go further. Now I am back at that same point in the cycle. I'm at y equals 1 and the slope is positive. And notice, I traveled. My change in x was the length of the period. It was 2.