It’s not everyday we can celebrate the birthday of a person who is both a pre-eminent mathematician and an Olympic medalist...
But one such was Harald Bohr (brother of much more famous physicist, Niels), b. April 22, 1887 (d. 1951) who worked in Mathematical analysis, founding the field of almost periodic functions, co-creating the Bohr–Landau theorem, and co-editing the journal Acta Mathematlca...
His sports career included winning a sliver medal for Denmark at the 1908 London Olympics in football. Denmark lost the final to Great Britain, but prior to that we creamed France A and France B, 17-1 and 9-0, respectively. Harald Bohr scored two goals against France B. The famous semifinal victory over France A included 10 goals by star striker Sophus “Curly Legs” Nielsen, and their defeat shocked the French so much that they refused to play in the bronze match...
Harald Bohr would have likely won a gold medal at the 1906 Intercalated Olympic Games in Athens where the Danish squad went undefeated, but he elected to stay at home to help his brother Niels prepare for his doctoral defense!
Photo: The 1908 Olympic football squad - Harald Bohr is the short guy in the back row, next to “Curly Legs” - the blond, short man in the middle...
A period of a function f : ℝ → X is a T ∈ ℝ such that for every t ∈ ℝ, we have f(t + T) = f(t). Note that X may be any set; the definition does not refer to any structure on X other than its equality relation.
For every function f : ℝ → X, the set of periods of f is an additive subgroup of ℝ:
For any two periods T1 and T2 of f and every t ∈ ℝ, we have
f(t + T1 + T2) = f(t + T1) = f(t),
so T1 + T2…
Hi, we're doing sinusoidal functions and I have no clue how to find the sinusoidal finctions for these tide/temperature charts. The question says "Find sinusoidal functions of the form y=AcosB(x+/-C)+/-D or y=AsinB(x+/-C)+/-D for both the tide and temperature data. Your data may not be symmetrical so look for an approximation. Show all work to find A, B, C, and D.
My data is in that google docs document, and I would appreciate it if you could answer this by tonight as I have to turn this in tomorrow, but if not I still appreciate the help! Your blog is amazing, and I love math I'm honestly really sad that it isn't taught the way it is supposed to, but it will always be my favorite subject, even if I have trouble understanding it!
This post should cover everything you'll need to determine a sinusoidal function in those forms. When it comes to determining data from graphs, just try to read key points as accurately possible and go from there! If you get stuck, come on back :)
So, periodic, or wave functions-- like sine, cosine, tangent, etc.-- are just like any other function in that you can "put a number in" and "get a number out". They're different, however, in that they repeat in cycles/periods. With word problems like this we just have to break it down to figure out the exact process that happens to the number we "put in". So what we want to figure out is
what values are being input
what values are being output
where the cycle starts and stops
any other relationships we can determine easily
and then build on that.
Here it's sort of obvious that our cycle is going to relate to a 24-hour period. There are, generally, two high tides and two low tides in every 24 hour day based on the Moon's orbit around us (because gravity is crazy! But let's ignore that for now). So, in actuality, one cycle will be close to twelve hours, after which point it repeats itself.
We are also told we want to "model the depth of the water x hours after midnight". What this means is that we want to be able to say "hey, it's ___ o'clock and therefore the water is ___ feet deep." Meaning, our x input is going to be "hours after midnight" and our output (y value on the graph) will be the depth of the water.
Next up: so just how to these two values relate? Well, we are told that this will be a wave function, specifically a cosine function in this case. That gives us a starting place for what kind of features we'll be looking for. If it were a straight line we'd be looking for intercepts, slopes, etc., but since it's a wave we're going to try to look for things like how high and low our wave values go (our output values, in this case, water depth), and how long it takes for these changes to occur. This is because these are sinusoidal waves-- based on circles. There are other waves, like square waves, triangles, and others, but we'll ignore those for now. Here's a handy gif for picturing how this relates to "circles":
The comparison of high and low values is going to be our amplitude. In general, amplitude is going to be the greatest magnitude of distance from the center/baseline-- in either direction. So, on a basic sinusoidal wave ( cos(x) or sin(x) ), our baseline is "y=0" and our amplitude is "1". The highest output value achieved with these functions is 1, and the lowest we can get is -1, so the furthest we will ever be from 0, with these curves, will be one unit away.
Building on top of that is the peak-to-peak amplitude. Just as we determined what our peak values will look like, now we're going to compare our highest value to our lowest. This is useful because if we don't know our "baseline", it will always be the midway point between our highest and lowest values. In order to find the peak-to-peak amplitude, we simply have to subtract our lowest value (the "trough") from our highest value (the "peak").
So, let's look at the amplitude of our example. We know high tide is 13ft at 3am and we know that low tide is 5ft at 9:30am. From this we can figure out our peak-to-peak amplitude, and from that get our baseline.
13ft-5ft=8ft, which is our peak-to-peak amplitude
8ft/2=4ft, which means at any point in the cycle the water will be no more than 4ft higher or 4ft lower that our base waterline. In order to figure the baseline we simply subtract the amplitude from the peak value, or add the amplitude to the trough.
13ft - 4ft = 9ft
5ft + 4ft = 9ft
(They are the same which means it checks out!)
So our "baseline" is the line y=9ft. Onwards!
Now we're going to use these values that we know to determine the cycle. But wait! "I already thought it's 12 hours?!", you're saying. Well, that was a rough guess, but we want to make sure our model is actually backed by math so we can make accurate predictions later on.
So, again, we know these two points, the peak and the trough. We already know that the "y" distance between them is the peak-to-peak amplitude, so now we will look at the "x" distance between them. This will be one half the length of a full cycle, as the function must then continue to its starting point to being the whole thing over again.
For the sake of writing functions, it will be best to turn our hours:minutes notation into simply hours, with minutes as fractions of hours. So our "9:30" low tide will become 9 1/2 or 9.5 hours after midnight.
Our peak occurs after 3 hours, and our trough occurs after 9.5, so again we simply find the difference:
9.5hrs-3hrs=6.5hrs
6.5hrs is then the REAL half cycle, and a full cycle will be every 13 hours. As it relates to our problem, this is because we determine a 24-hour day from Earth's axial rotation and how we see the sun, which doesn't exactly line up with the Moon's orbit around us, although that got us in the ballpark. The reason for this difference is that both the Earth is rotating AND the Moon is orbiting, not just one or the other. So after 24 hours the Earth is back to its same location in relationship to time of day (e.g. "midnight"), but the Moon has also rotated slightly around us, so in order to have the same location relationship between the moon and our original point takes a little bit more of time to catch up. You don't really have to worry about trying to picture this too much, although I will draw some helpful diagrams if anyone is curious. It's more just to make sure that you can understand how "word problems" relate math to reality (and why you should always check your work-- 13 is not the same as 12!).
GETTING BACK TO THE MATH! Now we know the period (13 hours), the amplitude (4ft), and our axis value around which the tides are oscillating (9ft). We have to take a basic cosine function and manipulate it to have these values.
A general cosine function looks like this:
y=Acos(B(x+C)) +D
A, B, C, and D represent the different things we can "do" to the function y=cos(x) to get it exactly where we want it. For y=cos(x), what you're actually saying is that A=1, B=1, and C and D both equal 0. So what do each one of these things do?
We'll start with the more straightforward bits, A and D. These two things both operate "outside" the cosine. That is, they affect our output only after our input has already "gone through" the cosine. A multiplies our cosine value, so it's going to stretch that number. In relationship to our graph what it's doing is stretching the amplitude. So what is our A value?
In y=cos(x), we know our amplitude is 1. We also know our A value is 1, which shows that there is a 1:1 relationship between the desired amplitude and A. We already know our amplitude to be 4, so that's what our A value will be as well!
So how about that D value? We know that it, too, operates on a value that's already been cosine'd. (Not co-signed!) However rather than multiplying or dividing each value, it changes each one consistently-- and this turns out to be our "axis" or "baseline" that we were referring to. When it comes to the graphing, this is usually called the "vertical shift", because it takes the basic y=cos(x) and shifts the whole thing up or down. In y=cos(x) we know that the "baseline" is y=0, which we also know is the D value. So again, D will be whatever we want our new "axis" to be-- in this case, we wanted it to be 9.
So right now, we're about halfway to writing our the function and it looks something like this:
y=4cos(B(x+C)) +9
This is where it gets a little bit tricky, because B and C both affect the input before it goes "into the cosine". (That sounds like a bad horror movie name.) B, like A, is going to stretch the function, but because it affects the number being input into the cosine, it will affect the horizontal axis. C, like D, will cause a shift, but it too will be the horizontal axis. So both B and C are going to relate to the period of the function and where it's "placed".
What we know going into this is that our period is 13 hours (our x units) and that our peak occurs at hour 3.
Remember how this all relates back to circles? Well our "input", or x (or hours), when relating to the circle, are actually the number of radians we've traveled going around and around and around. The length of one radian is that circle's radius, and we know from the circumference that a trip once around the circle will be 2π radians, or approximately 6.28 radius units, if it helps you to think about it that way. What else do we know about this? That "one trip" is also one period (one cycle!) on our wave function. How does this help us? Well, we can compare these values with the basic y=cos(x) function to our own function to figure out B.
The relationship is this:
Bp=2π
(π is the number pi, it kinda looks odd on tumblr's font.)
B is the frequency, or the "number of cycle segments" in a function compared to plain ol' cos(x), which is that 2π value. So in cos(x), there's only one cycle per 2π, and B=1.
p, then, is the period. This is the length of each segment in x units (so in our example, time). In cos(x), the period is 2π.
2 π *(1) = 2π, so it's easy to see how this is derived from cos(x). However it might seem a little more complicated when comparing it to your own function, but it's how we're going to find out what we need to know.
Just to recap:
B= # of segments, or frequency
p=period, or the length of each segment
Bp=2π
It's kind of like dealing with unit conversion in chemistry or physics, in that it is a relationship derived from a circle, so if you "balance your units" right, you will get a relationship to your circle.
So how can we solve this? Well, we know our period is 13 hours, and it's a given value from the function that Bp will always be 2 π, so that only leaves us with one unknown, B. Simply fill in the pieces and solve:
Bp=2π
B(13)=2π
B=2π/13 = approximately .483
All we have left to find is C, and we can go about it two ways, conceptually or plug-and-chug.
Conceptually we know that C will cause a horizontal shift. It's easiest to compare our first peaks between cos(x) and our own function. In cos(x) the first peak occurs at 0, but in our function that peak occurs at 3. This means we want to start our function (at 0) what cos(x) looks like "3 hours earlier", or a shift of -3.
With plug-and-chug, since all we have left to find is C, we could simply use the two points ((x,y) sets or hours-and-depth sets) to solve for C. However, it's best to understand it conceptually and then use the plug-and-chug method to check your work. If the point values work out, your function is all good to go! So let's check. Our function right now looks like this:
y=4cos(.483(x – 3)) + 9
Our first point on our graph is (3, 13) and our second is (9.5, 5), so let's check those and see if they work. BE CAREFUL! If you are using a calculator for this, you want to make sure that it is operating in RADIANS mode and not DEGREES. (I made this mistake doing this write-up, in fact.)
(3,13):
y=4cos(.483(3-3))+9
y=12.99... which is approximately 13
This works because we have done some rounding in our formula (2π/13, remember), and also because when working with numbers like pi and radians you may not always be getting exact values.
(9.5, 5):
y=4cos(.483(9.5)-3) + 9
y=5.000008 which is approximately 5
Looks like we’re all good in setting up the correct formula! The next part asks us to find out what depth the tide is at when it is 3pm. This is still going to be our x input value, but we can’t simply put in “3” because we know the tide won’t be the same at 3am as it is at 3pm—we already established that our cycle is actually 13 hours and not 12. So what are we actually asking for? Hours after midnight. If it helps, you can think about it as either sort of like a 24-hour clock (where 3 pm will be “15”), or counting forward from a fixed point. (Say, for example, that your graph started at “midnight on a Monday”. Another problem could ask “what is the depth of the tide at 6:45 on Tuesday?”, in which case you would add 6.75 hours onto the first 24 hours, or 30.75 hours counted forward in total.) So what you are going to do is plug “15” into your function as the x value:
f(x) = 4cos(.483(x-3)) + 9
f(15) = 4cos(.483(15-3))+9
(You should be able to get it from here, just make sure you’re still in radians mode!)
The graph you can either sort of freehand based on the different features we determined—period, peak, trough, amplitude, etc.—or chart a bunch of specific values. But now you know what it looks like!
(2000π/483 is about 13, which is where we determined our cycle begins again.)
So fun story, Hatteras was actually moved several hundred feet back shore a few years ago because of all the erosion. They used something like several thousand bars of Dove soap to grease some giant rails, put the whole lighthouse on it, and pushed it. And now it's a national civil engineering landmark because of it! :)
(And thanks to H2 for catching my formula error in my rough draft!)
