 Hi, everyone. So in this video, I would like to go ahead and go through some of what's going on in this final project. So let me go ahead and switch my screen over. And so what I have up here are just some clippings from the project handout. And really, I will go ahead and leave it up to all of you to sort of read through the full handout and see where these equations are coming from, what the scientific background is. This will be an important part of your write-up. But I trust that within your group so you can arrive at some understanding of what that situation is. And so today what I'll talk about instead is really just the mathematical derivation that's going to be happening in the middle, sort of what you have to solve for and how the equations might be set up. And I'll maybe say a little bit just about project structure as well. Just maybe how to go about laying things out. So here I just have part of the introduction section clipped here. And this is just coming from the project handout. And up here you can sort of read if you'd like sort of on your own time where things are coming from scientifically. What we're going to be most concerned about is this section 4, this physical situation. Because this is where we start seeing some of these mathematical equations. So the overall thing here is that we're trying to measure some sort of oscillatory behavior in these mitochondria. And so most oscillatory behavior in science can be measured by some combination of cosines and sines. So we'll make this sort of simplifying assumption that maybe it's just governed by one wave function. So maybe just a sine function. And we'll have a lot of parameters, some unknown numbers associated with this function. And maybe we'll go out into the real world, we'll measure some data, maybe take a few data points. And we'll try to use this data to somehow solve for all of these unknown parameters we're inserting into our function. So we can actually at the end of the day get a function just of two variables with no unknown parameters. And this is something we can plot on a graph and analyze in other ways. So the oscillation will depend on a lot of things. But at the end of the day we want to get it down to depending on just one thing, which is just the temperature. Or, well, okay. I should be a little more precise here. The oscillation is a function of time. So the oscillation really depends on time. This oscillation will have some sort of period associated to it. And that period won't depend on time, but that period will depend on temperature. And so it's a little bit, there's some sort of interdependence between some of these different variables that are showing up. So the two things to notice here are that we have this k is this sort of exponential function. And then we also have k appearing in this other function down here inside of the sine function we're using to model this oscillation. So I'll come back to that in a moment. I want to say, before we get too far into things, just maybe how to go about laying out the structure where this mathematical analysis will come into play. So I think what you want to probably do first is to have some sort of introduction or maybe some kind of abstract. This is a very common thing to have in sort of any scientific paper. It should probably be less than a page. Usually a couple of paragraphs is good. This is something you want your reader to be able to skim over to get some idea of, I mean maybe, so let me list out a few things. You want some overview, not really at a level of a lot of technical details or anything like that, but just if you're explaining this entire project to your friend or a family member and you only had 30 seconds or a minute to tell them what it was all about, that's the kind of stuff you would put in the abstract. So it would be kind of a high level overview. While you're doing that, you may also want to really state, anytime you're doing sort of any scientific endeavor or any mathematical endeavor, you're usually trying to answer some kind of question, right? And it's good to just state outright what is that question you're trying to answer. And so it actually may be useful to write this introduction or this abstract at the very end of your project, because the very last thing you do and use it as a way to kind of summarize everything you've done, figure out what question were you asking to begin with and what question did you actually answer. But it's good to actually state what's the whole point of the full report and what question are you trying to answer. And finally, you want to say something about the actual results, which is another reason that you may want to save the introduction for the end, because you won't know what your results are until you have them, of course. But this is, when I say results, I mean your results. So you are doing some kind of investigation into this sort of scientific situation that's laid out. You're going to be doing some mathematical analysis and maybe just in a couple of sentences, try to give the reader an overview of what conclusions did you reach, what analysis did you do and what did that reveal to you about the situation. Okay, so you'll have this sort of introductory section and that's its own thing, this abstract. You may be on another section where you talk about the physical situation. And this is the place to really go into detail about what's really happening with the situation. So if you go and look through the project handout, there's sort of a lot of details coming from this original paper that they're talking about. And you'd really want to explain enough of this detail to your reader that they see sort of what the underlying physical situation is and they have some good understanding of it. It's up to you to choose kind of what level of technical detail that needs to be. You might just go with the stuff that's coming from the project handout itself. You might go and try to look things up on Wikipedia or find some other primary or secondary sources to add some additional information to your report to help sort of better explain your results and your analysis. And one thing to keep in mind here is that sort of with any kind of report like this, you really want to be telling the reader some kind of story. There should be some kind of narrative to it. And hopefully, as you go to write the intro or the abstract, that narrative becomes more clear to you. And so what I would think about when I'm writing this physical situation section is, you know, how is it supporting your narrative? And so use that to guide how much technical detail to include. And then this is the one we'll be focusing on. There'll be some mathematical analysis. And this will be quite a large section. So maybe you want to break it up into several subsections. This is the one we'll talk about today. And so in this sort of physical situation section, you may want to start describing these equations, you know, what they're coming from, what their general forms are, maybe just a little bit of that. You can also do this in the first part of your mathematical analysis section to introduce these equations. This is where the majority of your work will go. Well, the mathematical side of the work anyways, as opposed to the writing side. And so, yeah, I'm just going to say there'll be sort of like a subsection one and a subsection two. And there will be, these are kind of laid out in the project handout. There are really two different kinds of analysis you have to do. One of them you're sort of given two data points. We'll talk about this one in depth and you need to go and solve for some parameters. And the second part of the analysis is really doing some of the first part again, except for you'll be working, you're kind of using some different sort of information to begin with. And then for the end, maybe you'll just want some conclusion. Well, I'm doubt that what I've heard about, you know, what makes a great paper, people often say, well, you should tell them what you're about to tell them. So, you know, that's your introduction. And then you tell them the thing that you mean to tell them. That's, you know, maybe this middle part of the physical situation in the analysis. And the conclusion is, you know, tell them what you told them. So, remind them of, you know, what were the key points of the sort of middle piece of your report. So, I mean, this can be like pretty similar to the abstract, the introduction. But maybe you want to be, you can kind of go into a little bit more detail in the conclusion. You can talk more about the analysis because you can assume your reader has already read that part and has some idea of what's going on now. And this is maybe a good place to, you know, to sort of restate kind of what the main question was, restate kind of what your conclusion about, the answer to that question, what that was. And to maybe think about, like, what additional questions does this analysis open up for you? You know, the usual kind of thing you have in a scientific paper is kind of like, what further work would you do in this direction? That's a good thing to include in your conclusion. So I've started this over here because these are the two sections we'll be looking at, at least in this video, the rest of it will be up to you and your group in terms of writing. And so let's just dive right into it. Let's look at the mathematical analysis. All right, so we started off with this, just coming from the handout. We have this sort of K, whatever this thing is. We're moving in a bit. And this K equals A E to the E over RT. So we're just given this. And if you read the handout, yeah, so the first thing here, and I think this is kind of important to keep track of, is what is supposed to be our independent variable at the end of the day? So in this bit, we're trying to find a period as a function of temperature. So we want temperature to end up being the independent variable. And then we should also identify kind of what things are our constants, or what things do we have to solve for. So maybe I'll just do, in this color, this green-blue, this R constant, because this is something that's seemingly like an unknown number or parameter, but we actually have a value for it. And it's something like R is equal to 8.314. And I don't know if this is exact, so maybe I'll indicate to you and to my readers, as I'm writing this, that maybe this is an approximation, unless I know that this is the precise exact value. So I have that, which means that there are only really two things left to solve for. I really want to know what this A is and what this E is. And if you kind of look at the second part of this situation, you have some kind of function like I'm just going to write it as f of t, depending on time. And it will be this sine function. Again, this is just coming directly from the handout at this point. We get something like 2.6 times t over square root of k, plus some parameter phi. And sort of like we saw in class, the first thing I would do is notice that this is not in standard form. So before I start any analysis at all, maybe I just want to go ahead and put this into standard form, so I don't confuse myself later. So let's see how it goes. So what do I need to do when I'm just going to write the same thing out? There's kind of a trick to it. Whatever I see in front of the, so let me actually go back here and label my independent variable, t, showing up there. So the trick is that I factor everything out of, everything in front of t, the coefficient in front of t, I want to factor that out from everything else. So I'm just going to literally write 2.6 over square root of k. The only difference is I have to put a left opening for it to see, for in there. Okay, I need to maybe put plus or minus something here, and then you can close that parentheses and close that. All of this is still the argument of the sine function. Maybe for standard form, right there's this minus that shows up. So maybe I want to have this be minus something. Maybe I'll have a minus five there. And I just have to check whatever I do here. I need to be able to distribute this back in and get exactly what I'm seeing up here. So the trick is to take this thing here and put it in the denominator of the second term. So 2.6 divided by square root of k. And then you can just check, right, if I multiply this and this, I get precisely this term showing up there. And if on the other hand I multiply this, the second term, what happens is that this thing exactly cancels what's happening in the denominator over there. Essentially, I set it up exactly so it would do that thing. And then I just wind up with whatever's in the numerator. So after I have this cancel the denominator then I'm just left with minus five, which is, oh good, indeed a plus five. Okay, so I would leave this form of the equation for now. And this is just so whatever number you end up getting here, you know that that's going to be a shift to the right. It's a little bit tricky because there's a negative sign in there, but maybe this phi ends up being negative already. So if this phi is like negative two or something, then maybe this thing is a positive two in the numerator. And if this whole thing is positive then you're shifting something off to the right by whatever that is. And really if you think about this, what I'm saying here is that this thing is actually the phase shift phi from the standard form we saw in class. And this thing out in front is really the frequency of omega. And so the first thing, if you kind of go through this part of the project, is the first thing to look at is that we'll be looking at the period. And by the way, just notice that for now I'm keeping this k around even though we have another name for k. Let me tell you what it is. So here's one name for k. It's just k itself. And here's another name for k which is all of this stuff. The k is on the left-hand side, all this stuff on the right-hand side. I could sub in for k at any point if I want to. Which I may want to do here in a moment. So one thing I can do, I know that the period is given by just the following formula. You just remember for the standard form really the normal side of the normal cosine function at a period of 2 pi. And the frequency just has an effect while it's supposed to be like some kind of horizontal dilation or some expansion or some contraction in the horizontal direction. So this is just saying that I just need to scale the standard period by whatever this thing I'm using to scale horizontally to get the new period. I'm scaling the period down in this case. Okay, but I have a name for omega. So I'm just going to write that out. Omega was literally all of this stuff. So I will put that in the denominator. And maybe just to make things a little bit easier to handle I'll not be dividing by something that's already a fraction. Maybe I'll just make it all one fraction by moving the square root of K upstairs. Like it's something like this upstairs, 2 pi square root of K. Just the 2.6 downstairs. And this thing was supposed to be able to P. And I guess right now P is really a, it's a function of K, right? Also appearing here is a number. 2 pi 2.6. There's this other kind of unknown number that it sort of depends on. K. So I guess what I want to say here is now we should try to expand K out. You have that other name for it coming from earlier. This one here. So I need to take this entire thing. Everywhere I saw K, I put this thing now. So let me just put that inside of the square root. I have a 2 pi out in front. Then I have some fraction where 2.6 is in the denominator. And just to clean things up a little bit, I'm just going to pull the 2.6 out and have this 2 pi over 2.6 times A to the capital E over RT. And then I take the square root, that entire thing. Maybe just a note to myself, R is known. We have a numerical value for that. And then we also want a T to still be our independent variable here. And so we kind of want this P to just depend on T. But right now it depends on a bunch of other stuff. It depends on this A that's showing up. And it depends on this E up there. And I think that's it. Because this lower case E is just always constant, 2.7-ish, whatever it is, I would look it up. And then we have this R, some numerical value for it. And then everything here is a number. So this period is something that depends on multiple variables. These three variables. And so what we want to do is kind of change our perspective and say we don't want these to be variables. We're going to go out and measure stuff in the real world and make these more like unknown numbers or parameters. And so we're going to try and solve for it. So we can't handle a function with three variables. Just using the techniques from this class. This is some higher dimensional kind of object. It's really hard to graph it on a two dimensional piece of paper or even like in 3D graphing software. This is something I guess it's four dimensional. So it's a little bit too hard to work with. So we want to get it back down to something we're used to seeing in this class, which is where we have essentially one variable for the function, one independent variable, instead of the three that we have here. So that's the name of the game. It's to try to solve for these two. So we can make it a function just of capital T for the temperature. Okay, so let me just write that out. So we want P of T. So we want to solve for actual numerical values of A and B. And so if you start reading a little bit below where these equations are laid out, you see that we're actually given, so we go out into the world. We use our experimental apparatus and we measure the period of these oscillations and we measure the temperature associated with that. So we get an ordered pair, let's say T0 and P0. So we're just measuring the temperature and then we're measuring the period and we get some actual numerical values for these. So I think this ends up being like 25 degrees Celsius or something and then maybe 121 something. You have to check what the actual... By the way, there are units for all of these things, so you guys have to keep track of those. So here I'm just not including any of them, but for the actual write-up, it'll be important. So you measure one data point like this and then I don't know, you wait some time and then you measure some different data point, T1, P1. And this ends up being 37 and 104. And we're doing this because we kind of want to think of these as like inputs and outputs or something. So we're measuring kind of an input T, it's like an input temperature, and we're assuming that the oscillations are a function of T. So they're kind of like the outputs associated with that. And so we're hoping we can use... We have two things we need to solve for and we've taken two measurements. So if we can get two equations and two unknowns, this is something hopefully dual. So let's see what happens. By dual, I mean you can hopefully solve this kind of thing. Two equations and two unknowns. If you just had one equation and two unknowns, that wouldn't be enough equations to solve for everything. And I guess if you had more unknowns than equations, then that would still be fine, I guess. So let me just try to translate what this actually means when we have these two data points. Essentially what it means is that you want to take this T-naught, which is 25, and you want to go ahead and plug it into this equation you're seeing up here. So the T is going to go in that slot there. And this P-naught is telling you what the output is. So that's what's going to be showing up on that, the left-hand side of your equation. And so then you're going to write an equation doing that by plugging in these two. You just do the same thing of writing an equation using these bottom two. Where this T-1, again, you'll just plug in as the T in this equation up here. And this P-1 will be everything appearing on the left-hand side. So let me just write down what I said there. So the first one is the output is 121. So that's what's showing up on the left-hand side. And the input was 25. And so let me save myself some work here. I'll just go back up, copy this entire equation, and we will substitute in A value. So it's equal to this entire thing, except for these T's. These are like my independent variables, T. The first one we're doing is this T-naught. Then I have a numerical value for T-naught, which would be something like 25 if you wanted to plug that in. And now I just played the exact same game with my second data point. So I'm just going to make a literal copy of this, except for the output is different now. The output is 104. So let me change that. And the input is now 37. So let me change that. And let me just be a little bit careful with my notation here. This should be like r times 25. And that whole thing is in the denominator, and then all of this stuff is still up in the exponent when I'm exponentiating. Same thing here. I have r times 37, or r times 37. r times 37 is in the denominator, but then this whole thing is being exponentiated. So these are our two equations in two unknowns. And that means you can solve for A and those of that. Just the parameters that are showing up there. And there. Okay, there's a little bit of a trick to this. And it's actually a trick we've seen before. When I think this came up when we were doing some of these problems about like having modeling bacteria in like a petri dish or something, like they're weighed over time. So I guess this was around the end of unit two. And we had equations that were sort of similar to this. One trick we can do is kind of divide both equations, divide both sides. And that'll be the way we get off the ground with this one. So what I'm actually going to do is do a sort of a general form of this just because it's up to you guys to actually work through the derivation with the actual numbers in it. So I'll try to do a general form, which should give you sort of a nice recipe that you can follow. So instead of having those numbers, I'm just going to leave them. Well, let me actually do this. So I'll take this entire thing, maybe more clear to see where everything is coming from. So this was like my equation one. This is my equation two. And what I'm doing here is we'll just do a slightly more general version of these two equations where we don't use numbers. I'm just going to use P naught there. And this whole number, I'll just give it a name to keep things simple. Let me just call it C. And then this thing up here, this was a purple T naught. So everything's just going to be a parameter here. You should be able to plug in the actual parameters you get and follow roughly the same process. Well, exactly the same process as it turns out. But you'll be able to plug in numbers to check. My equation two, the only difference is this was a P1 and this was a T1. And I'll show you how to solve for the A and the E parameter in this general setting. So the trick is to take equation one and divide it by equation two. And so on the left-hand side, you'll get P naught over P1. On the right-hand side, so if you just kind of look at what's happening here, just to save myself some writing, when I take this whole thing and divide it by this, it's going to be nice is that this, whatever this constant was out in front is going to cancel. This is just multiplying both things. When I divide it, I can just pull it out and it'll become a one. So I'm left with this whole thing divided by this whole thing. In fact, maybe one nice way to see it is just to put a line here or this ratio is equal to this ratio. So I'll be able to cancel these guys. And I'll be left with just this stuff. And I'll have a ratio of square roots, which means I can put the whole thing under a square root. So let me just start from there. On the right-hand side, I'm going to have some giant square root. And the thing I'm going to have upstairs is exactly this. So let me just copy it. And the thing I'll have downstairs is the radicand here. The fraction will be occurring inside the square root. And we see that even nicer stuff is happening. So I'm getting some cancellation of the a's. And really, that's the whole point of this maneuver is that you want to eliminate one variable. Let me just kind of spell out what happens here after we cancel some stuff. So the first thing is that we can divide the a's out. Get something like that. The second thing is that I know how to divide exponentials. If I have an exponential downstairs, I can move it upstairs. But there's a cost. I have to pay the price. And the price is that I have to change the sign on the exponential to a negative. So I can multiply these upstairs as long as I change the sign of what I'm exponentiating to a negative. And I can simplify this even more. I guess we're going to a new page. There's another exponent property I can use that if I have like bases, then multiplication downstairs and the fact that these are like bases can be changed into addition upstairs, which means that I can delete this, this, this, and this and change this to a plus, all of this stuff. And so let me just write this out a little bit more clearly. So on the left-hand side, we still just have this ratio, p0 over p1. On the right-hand side, we're going to have a square root of some stuff. This stuff is going to be e, so little e to the capital E over r times t0 in the denominator, plus capital E over r times t1 in the denominator. All of that is in the exponent of this, of what I'm exponentiating here, e to something. And I take the square root of all of this. And by the way, the thing that we're trying to solve for, well, e was one of them. A was another one, but A disappeared here. So we were able to cancel and A. And this is good in the short term because we want to solve for one of the variables. So it's nice that we only have one left. It's not great in the long term because we still want to find out what A is. But the point is that once we get e, it shouldn't be too bad to go back and find A. So what I'm going to do here is try to solve for e. I guess I should double-check this p0 over p1. This is just a number. This was given to us as some of the measured data. These were the output periods. This r was just a constant, this 8.314 joules or something like that. And this t0 and this t1 were also part of that data we measured. Remember that we measured a p0 and a t0 and then a p1 and t1. So we'll actually have numbers for these. So that really means that e is the only thing unknown in this whole picture that we have up. But okay, now it's just an algebra to isolate that e to free it from what's happening in there. So let me go ahead and square both sides of this and I can remove this square root on the right-hand side. I have to do something here. So I have e's in the exponent. So if I want to move those downstairs, I need to take the natural log of both sides. So let me do that. I guess really remember that I could take any log of both sides. All logs have this property. So I could take like a log base alpha like we did in unit 2. But maybe I see matching it up with the base of log with the base of what I'm exponentiating is always a nice thing. So I'll make it log base e. So I need to take the natural log of the left-hand side and all things being equal. I have to also do that to the right-hand side if I want to maintain this equality. Okay, so I have this whole thing. Since I'm looking in this set of square brackets on the right-hand side, I'm looking inside the log. Everything in the square brackets is the argument of the log. And I'm seeing that the entire argument is all raised to a power, which means that I can pull that power out. So I can take this and surgically extract that. I'll come back to it in a second. This log base e survives. And then this whole thing that was in the exponent gets pulled out in front. And I need to multiply those. But I'm being a little bit ridiculous, right? The natural log of e, I'm using funny square brackets, but it's really the same thing that we've seen in class before. The natural log of e is really just the log base e of e. And this is actually a question, what power do I raise e to to get the thing I see in the argument, which is e to the, well, to the 1. So this whole thing is 1. So I can just go ahead, I'm just multiplying in a 1 so I can just raise that. Okay, so I think I'm almost done. I think the next thing to do here would be to pull out the e. So I think I want to factor an e out of this first term. I'm left with a 1 and a 1 like that. I guess I have to factor it out of both terms here. And I would just double check, anytime I do a step like this, can I run it backwards? If I multiply this e in, I need to distribute it to both terms. And oh good, that gives me exactly what I had before. So I know this is a valid step. And just rewrite it slightly. And I think it's the last step. I think I just want to divide by everything that's not e. So just rearranging things a little bit. I have this natural log of stuff. And then I have all this other stuff that's not e. So let me just divide both sides by it. Not quite what I wanted. There we go. And I'm left with e equals all of this stuff. So these were all logical steps. So maybe I want to make sure I'm telling my reader what the flow of the logic was. We were solving some big long-term equations and it's nice if I can lay out a bunch of steps where in each step I'm doing kind of one simple operation, hopefully simple. So that way the reader can go back and trace through this computation themselves and verify that it actually works. It's kind of a long computation, but once you get to the end, you have e equals some stuff and it's mostly bookkeeping. The tools you're applying are, when we exponentiated some stuff, we took a log somewhere. Yeah, where do we do that? We did this part here. And so here we took a log. Here we used some exponential properties, like the fact that the natural log of e was equal to one. Stuff that was happening here was really just algebra. It's kind of complicated algebra. And here, you know, just more algebra. So I think it's worth seeing, like where was the actual hard part? Here, what did we do in this step? Here, I think we just used this property that e to the a... Did we do anything here? I think we just rewrote. This was just to clean things up a little bit. I think it was here going from this step to the next one. We used the fact that e to the a times e to the b is equal to e to the a plus b. This wasn't something special about e, though. This could have been any base alpha, just a property about exponents work. Here, we said that, yeah, there's some cancelling of a that happened. Here, we just did this sort of division step. So we've left our reader, you know, sort of a set of breadcrumbs, you know, a process, a recipe, something that they can walk through themselves. And so if I were writing my report, I would probably include a derivation like this to show the reader where this solution is coming from. But at the end, I got e equals something, and that's one of the things I was trying to solve for. So I would just put it there. And then I just note that here's something that's a number that's given to me if I go all the way back up. So we're in the weeds now, what were we originally given. We had t0, p0, t1, and p1, and these were all actually numbers. So if I go all the way back to the end, I see that I know p0 and p1. So I can just plug those into my calculator, take the ratio. I can square that. I can take the natural log of what I get, and that'll be some number. And then maybe separately, okay, I know this r was like an eight point something, and I know this t0 and t1. So I can plug all this stuff into my calculator, and I can divide these two guys. I'm gonna get some number for e. So there's nothing unknown left on the left-hand side. All just the numbers at this point. Okay, and now, so you've solved for e, and now you want to solve for a. So you just have to note that we can go back to, we had two equations. Let's see, all the way back up here. And I think these two are where we started. So let me just take, it doesn't really matter which one. Let me just take the first one. I go down here and paste that in. So what I can use is that I now know the value of e. So let me just make that. So there's no color on that. So t0 was an independent variable. So we'll know that. We'll have a numerical value for that. e is something we just found. We have a number for that. We have a number for r. And we know the p0 as well. That was one of the data points we measured, the p0 and the t0. Which means that a is the only unknown in this equation now, after I plugged in my number e. Which just means if I can solve this equation for a, then I think I'm home-free. So let's try that. So I just need to divide over this c first, and let me maybe do a couple of steps at once here. The first thing I did was divide through this c. The second thing I'm going to do is square both sides. And then I'm left with this radicand on the other side. And this is a little bit tricky, but just remember that the thing we're trying to solve for is a. So whatever this thing is on the right, it's just some number at the end of the day. So that's something I can divide over. I'm going to take this whole equation, make a fraction on the right-hand side and just divide this whole thing over. And that whole thing is equal to a. And maybe I'll just clean this up a tiny bit. Just remember that this whole thing being downstairs, I can move that upstairs at the cost of changing the sign on the exponent. So it's e to the negative capital e over r times t0, multiplied now by p0 over c squared. And these are all things that we know on the right-hand side. I guess this c was a little bit mysterious, but let me just point out where that was coming from. I go all the way back up here. If you remember, we kind of made up this name for c, but that was just because I didn't want to write 2 pi over 2.6 everywhere. So this is actually a number that we know. I can plug this into my calculator. So if I go all the way back down, I'm seeing that everything on the right-hand side, the c I know, p0, I can square it. I know the t0 because it came from the t0, p0 measured pair. I know r that's given to me. And then e was the thing I found from that long first step. So all of this stuff is a number. And then I can just plug this into my calculator and multiply by this, and I'll get an actual number for a. Okay, so I think that is all I want to say about that. Right, so that's at least how you solve for these two parameters. So let's go ahead and take a look really quickly at sort of what the next part is. So you notice that this part didn't really use anything about this Q10 business that's showing up. So this is all really about sort of part two of the analysis. What we've been doing has been everything in this paragraph. So I'll just have these two notes and then I'll upload these so you guys have a version available. So this is asking us right from the information for the oscillations can be attained as a period as a function of time. So we want to determine what this function of time is. And we should say something about its important features. Alright, so the periods. Well, okay, this is the typo. I guess I should mention this should say temperature period as a function of temperature. And if I go kind of all the way back up to where I started with this whole problem, we really had this kind of thing that was sort of our original model and it depended on all of these other parameters, but then we solved for a bunch of the parameters. So I can get rid of this A and this E that we were trying to solve for because we have numbers for them now. And we have a function like this, which is our period as a function of temperature. And this A was something we have an actual number for. This E is now something we have an actual number for. So now we have a function that we can just go plot. And so this is I would maybe try putting this into Desmos and just answering some of these questions about like what is the behavior of this function? What kind of function is it? What's its limiting behavior as say the temperature goes off to infinity. The temperature approaches zero. So it wants you to potentially graph this. You might ask yourself is this a trigonometric function or is this more like an exponential function? If so, kind of discuss its properties in either way. Talk about whether it's like exponential growth or exponential decay and sort of justify that. And then, yeah, talk about what happens at large and small temperatures. So you'll get a graph when you're doing this, by the way. It's a little bit tricky, but you're not doing axis as a function of y. On this axis, you should have temperature as your independent variable. And on this axis, you should have some values of p. And you're going to get some kind of, let's just draw some weird function. Who knows what it is? It'll be some p of t. You get from plugging in this entire equation into say a graphing calculator. So this would this would be all of sort of part one of the analysis. Just to remind you if we go all the way back up. That is essentially corresponding to this bit here. And now there's a sort of second part to the analysis that shows up in the handout. So that's coming from the second paragraph here. And I'll just give you a hint of how to get started on this one. Okay, so just to read what's happening happening here. It's telling you now that, okay, in this situation, so it's a new situation now. Instead of, I mean, you might imagine that somehow you're not able to measure the period in the temperature twice. So instead, what you're able to do is measure some t not p not. And see, they're saying there's a, there's 25 degrees Celsius and 118 seconds. So 25 degrees and 118. So that's kind of one data point you're able to take. And then the other data point you're able to take is something totally different. What you get is this 33 degrees of a temperature and then you get something called a temperature coefficient of 0.91. So instead you measure a t1 and maybe a q1. So the t1 is this first temperature, this temperature 33, and this q1 is 0.91. And so it's not really the same situation as we had before because we haven't measured the period here. We've measured some other coefficient, some other experimental device or something. And so this is like something we've talked about in mathematics before, sort of in class. One really nice thing is like we've already solved kind of a hard problem. Do we have to do any extra work? And so what I would really try to do here is try to take this and somehow reduce it to the problem that we already solved. So that's at least how I would approach this. The way I do that is by going and looking, so here's this temperature coefficient they're talking about. It's q10. I think I'm just going to call it q probably. Yeah, so it's a little bit confusing. Let's leave the 10 out for now. Just call that q. And so what they're saying is it's, okay, I guess they're using tau instead of t. But what they're saying is that q is slightly larger here. q is equal to tau1 over tau2. And then they raise it to the exponent 10 over t1 minus t2. Sorry, I think I just said a minute ago that they'd written tau's for t's, but really these are two different things. So tau's I think are the coefficients. Here, let's see. I'm going to go back to this. tau1 is the period of the oscillations at the temperature t1. So I think what this really means is that what we've been calling p is exactly what they're calling tau in the handout. So just the periods of these oscillations. And so what's happening here is that we're measuring, we have, let's see, I guess they're doing t1 and t2 as well. So let's do it like this. We'll do a t1 and a p1 up here. And then let's do a t2 and a q2 there. And so the information that we have is the following. So there's a t1 appearing here. And then we've also measured a t1. This is 25. There's a p1 that we've measured, this 118. So that's showing up there. Maybe I'll just highlight them as we go. There's a t1 and a p1 corresponding to that t1 and that p1. We have actual numbers for them. And then we have a t2, which is a 33. So there's that. And then we also have this value of q. And so we're really kind of measuring this output. And that means that the thing we need to solve for is maybe this p2. And if we can do that, then we can use the fact that, you know, let's say we get that p2. Then we'll have a data point of the form t2, p2. And now I can just kind of use these two data points. So now I have a t and a p and then another t and a p. So I've measured it twice. And I'm really just back in the same first situation if you think about it. Let's just go back up here. This was sort of the original setup here is that we had these two data points in the form. So I think if I can just solve for that p2 and get another data point, then I can probably just run that exact same argument I did at first. And in fact, I've already solved it. So I can probably just start plugging stuff in. And okay, so I guess what that's saying here, I'll just work out a little bit of a corium. So if you have this like q2, I guess is what I'm calling it. It's equal to p1 over p2 to the 10 over t1 minus t2. Just keep in mind that the thing you're trying to solve for is p2. What I would do is just, I guess you could take the raise both sides to the 1 over this exponent power if you wanted to. Maybe something else equally easy is just to apply along to both sides. So we'll just do a log base alpha q2 is equal to a log base alpha of the entire right hand side. And I can take this and I see the entire argument of a log is being exponentiated. So that's the best situation, the situation where I can move the exponent out, something like that. And just to save myself, I will do it in another step. I'll go ahead and divide over this whole thing. And I have a fraction like that appearing on the left equal to all of this stuff on the right. And I've actually changed my mind. I think this actually ends up more difficult than what I originally suggested. So let's not do this. Instead, let's start with this one here and go with what I said to begin with. That seems like the easiest way to go. Or you raise both sides to the reciprocal of this power I'm seeing on the right hand side. What I do is I raise this side to the 1 over exactly this thing, whatever that power. And I just need to do the same thing to the other side. I raise it to that power. And the whole point of this maneuver is that the right hand side simplifies considerably. These two things. It's like I'm raising A to the B to the 1 over B. Sorry, let me put these down here. Doing something that's like A to the B to the 1 over B. And exponent rules say that if I exponentiate twice, then I multiply. So it's A to the B times 1 over B, which is just A to the 1. So it's kind of a way to cancel stuff from the exponent like that. And that just means that I can take this whole thing here. Essentially this whole business here is canceling this here. Let me just get rid of both of those. And I'm left with this. And for my last step, I would just multiply the P2 to the right hand side or the left hand side. And I would have a P1 left on the right hand side. And then I will multiply through by the negative of this to get something like that. So I did a couple of things there. Really what I did is swapped this P2 with this Q2. It's sort of a cross multiplication thing here. If you want, I did it in two steps. I multiplied through by P2 to both sides. So then I had nothing down here and then a P2 and a P2. Well, P2 times P2, so that canceled these. And then after I was done with that, I divided both sides through by this. Which is the same as multiplying by the negative... Well, same thing where I put a negative sign on the exponent. So I'll leave it up to you to work out how this actually works. But you can go ahead and solve for P2 from that first equation. It's just a little bit of algebra and some stuff from unit 2. And then once you have P2, we go back up here. This was something you were solving for. And now you can just run that same analysis from part one again. So I think that's all I really wanted to say about how to work through the mathematical derivation here. I will probably hold some student hours at some point if people have other questions. But hopefully people get a chance to watch this video before they attend those. And then also to watch this before they actually do their write-ups. So best of luck and feel free to reach out the email if you have any other questions.