 We are at lecture 25. We are once again in the middle of a problem. This is a exponential decay problem. So let's reread the problem, get it to the point where we had it, and then try to get a solution from there, see if it's a realistic answer. The location of the topic in the book, we are in chapter 7, section 4. So we've got a couple of topics yet to cover here. This one is a problem that we started yesterday on carbon dating, trying to determine how old something is based on the amount of C14 that remains in that formerly living thing, whether it's a plant or an animal. The radioactive element carbon 14 has a half-life of 5,750 years. That's probably why it's pretty good to use to date something that we think is pretty old. The percentage of C14 present in the remains of plants and animals can be used to determine age. Now this specific problem, archaeologists found that the linen wrapping from one of the Dead Sea Scrolls had lost 22.3% of its carbon 14. How old was the linen wrapping? So what we did first yesterday, we don't really know how much we start with. It doesn't really matter as long as you end up with half of that in that timeframe, which is 5,750 years. Reduce both sides by Y sub zero. Solve for K by taking the natural log of both sides, and we got approximately this value. And again, I don't know if that's enough accuracy. It's probably more decimal places than we usually use, but in some problems where you're trying to determine if something is very, very old, you may want to keep more decimal places than that. We did get further than that. Not too much further. So now we're back to the original equation. Now we know what K is for this particular stuff, C14. The statement says one of the linen wrappings from the Dead Sea Scrolls had lost 22.3% of its carbon 14. So we're assuming that it started with 100%, and if it lost 22.3%, it has 77.7% remaining. So 100% is one. Right, that is a decimal. That's where we leave off. So we want to solve for T, which is how old is this linen wrapping? Questions to that point. Not a good day to wear that sweatshirt. I don't like that sweatshirt today. I think you need to give it back. We got beat by Carolina last night, and here's a Carolina sweatshirt. It's a good thing it's not a task. I'd have to take points off. But we played pretty well, I thought, except for about four minutes in the second half. All right, what needs to happen right here? Natural log of both sides. The natural log of E to a power is that power, which that's what kind of helps the cause here is it gets the variable out of the exponent position and brings it back down where we can kind of algebraically deal with it. Divide by that coefficient, which happens to be the K value for C14. So I guess we'll find out if this is a realistic answer point. So it's about 2,093 years old. Realistic answer? Probably pretty realistic answer for this particular situation. Based on when they have tried to figure out based on what was written in the Dead Sea Scrolls, when it probably was written, and then when this says that it was written, probably based on how old the linen wrappings were, that's probably a fairly realistic problem, and that was years. Questions on that problem? Okay, let's continue with different types of modeling situations that involve exponential growth and decay. This is a little different order than the book, but I think it probably makes more sense to go to this example. If we were doing regular compound interest, the future amount or the final amount would be the principal amount. I don't know how familiar that looks, but hopefully it's fairly familiar. This is compound interest when we know the compounding period. It's annually, semi-annually, monthly, daily, hourly every minute. If you can put a number to N, or N is the number of compounding periods, we'll assume it per year. If it's annually, semi-annually, quarterly, monthly, what do you think banks use for daily? Interest compounded daily. We would like for them to use 365. They use 360. If we can get a specific number for N, then we can plug it into this formula, put it in this position, put it in this position, basically pull the calculator out and do the calculations. The problem is what happens when N is approaching infinity, so it's more than every day, more than every hour, more than every minute, more than every second. It's being compounded continuously. So that switches. We cannot use this model. We have to derive our own version of this model for continuously compounded interest. Now, before we get there, I just wrote that down just to show that this particular model will not suffice for N approaching infinity, but let's see with some data what it looks like when it's compounded. Let's just figure out a few values and then kind of compare that to what it looks like when it's compounded continuously. So let's say that the initial investment, the principal is 10,000. Let's give it a realistic interest rate. What's a realistic interest rate right now? Three? Three and a half maybe? We're real lucky. We'll have to check that. It might change tomorrow. Let's say time, let's say five years, and we'll take different N values. So if you have a calculator and you can help, this will make this go a little bit quicker. So let's see what happens when N is four. So we're going to compound quarterly, let's say compound monthly, and then we'll switch horses to the new model, and then we'll see what happens when N is compounded, when N is approaching infinity and it's compounded continuously. So do we talk about grand in here? Why that's called 10 grand? That never came up in here? That'd be a good little research topic for tomorrow. Somebody look up why 10,000 is called 10 grand. You probably won't believe it. I didn't believe it when I read it the first time. Can you just tell us? No. So why is it called grand? Very, very bizarre. Alright, so one plus the interest rate, and we're compounding quarterly. So N is four and T is five. Let's break down a little bit of the numbers that exist here in this formula before we go any further. 0.035 divided by four, wouldn't that be your interest rate per period? Is that correct? And if you add one to it, that tells us that they're going to give us all the money that we put in plus that interest rate for that quarter, right? So tell me what four times five is for this problem besides that it's 20. I know it's 20. No, it's not a trick question. Okay, so it's the total number of compounding periods. So that's the number of times you actually figure out the interest so that that interest that you just earned can make interest for you on the next period. So there's 20 compounding periods. This is the interest rate per period. You add one to it because you get everything you put in plus the additional amount. So the formula really kind of makes sense. So when N is four, what do we get? Did you run that through, Katie? 10,000? 11,000. 11903.4. 30? 03. Sorry. Thank you. 0.4. Yeah. We've already talked about that banks don't give you the extra if it's 0.4. I mean, they'll give you the penny, but if it's like that next decimal place is eight, you're not getting that. They keep that. So what is it? 399. 399? Who is it? 999 repeating? No. Oh, okay. Sorry to get excited there. That was 999 repeating. So 399, what do you think the bank's going to give you? 40? 30. No. 39. All right. Let's see what happens when N is 12. So now we're compounding monthly. Again, this extra inside of parentheses, 0.035 divided by 12 is your interest rate per month. And we will go through 60 compounding periods, 12 each year for five years. What is that? 4-2. So it's advantageous to compound monthly as opposed to compounding quarterly, even at this low interest rate for really not a long period of time, five years. Are we getting a consensus as anybody else running that through? Not that I distrust Katie. We're getting the same. We're getting some agreement. Okay. Now let's see what happens when N goes to infinity. So this is continuously compounded interest. So we can't use that model because how in the world do you put in N equals infinity into the model? So you're dividing by infinity and you're raising something to the infinity power. So we've got to kind of see what happens to the model before we... So an infinite number of compounding periods for this particular situation just to kind of help visualize what's actually going on. Let's make this substitution. And as long as we can justify everything, I think what happens to the K's, what the R is fixed, right? R is the interest rate, so it's a decimal number. And if N is approaching infinity and R is a fixed number, doesn't that mean that K has to get infinitely large for the left side and the right side to be the same? So if N is approaching infinity and if I'm going to make this substitution, you'll see the reason for the substitution, then it's also going to be true that this K value that we're using is also going to approach infinity. No way the equation could be maintained if that were not the case. So everywhere we see an N, we're going to put in a K times R. Any obvious reductions? R over R. So at this point we have that and I'm going to put some parenthesis in here. You tell me if you think that's legal. So I'm going to raise that to the K and then I'm going to raise that result to the RT. Is that equivalent to the statement above? Right? Because if you're raising something to the K and then turning around and raising that to the RT, wouldn't you multiply the exponents? Which is what we had on the previous step. So what's the advantage of taking 1 plus 1 over K and raising that to the K as K approaches infinity? I don't know, does that look familiar? This stuff that's in these brackets. Does that look familiar? I think that the answer there is no, that it does not look familiar. All right, well let's take that off to the side. 1 plus 1 over K to the K as K gets infinitely large. Let's let K be 100. That's not infinitely large yet, but it's a starting point. So what do we get for 1 plus 1 over 100 raised to the 100th power? This is kind of counterintuitive as well because you think, well, 1 100th is not very much. Isn't it E? Eventually it's going to be E. What do we get with 1 plus 1 over 100 to the 100th? 2.704. Let's do the same thing with K equal to 10,000. Are we, is there like a train outside the room? Just about to enter the room? Did we just kick on the air or something? Okay, tell me if I need to move out of the way for the train to come through. So if you don't believe that this thing is really approaching E, take K equals 10,000, which is still not nearly infinitely large, but it's pretty large. What's 1 plus 1 over 100 raised to the 100th power? 2.708. That's pretty close to the value E. And if you let K take on larger and larger and larger values, this quantity inside here is really just the number E. So it's getting a little cluttered there. If this part inside, inside the brackets, is really approaching the number E and you put that in for those brackets, you get an old friend, right? Or at least an acquaintance. So this is the model for continuously compounded interest. Notice there is not an N in there, right? We've dispensed with the N in the model by kind of renaming some things. We first of all called N K times R, made a little reduction in the final version. There's not a K times R in the problem, therefore there's not an N in the problem. Now isn't this the same model as this? Exactly the same, right? This is the future amount, that's what that is. This is the initial amount, that's what P is. It's the principal amount of money. This is the growth constant or decay constant. In this case it's a growth constant. That's exactly what interest rate is. And then obviously time is the same. So it's the same mathematical model we had before. So something about this particular growth situation says that the rate of growth is directly proportional to the amount of stuff that's present to begin with. So money in this case, the rate of growth of money is directly proportional to the amount of money that you have at any point in time. So back to this particular model, with this situation. P was 10 grand, R was three and a half percent, and time was five years. So I've never seen, and if you ever see this please bring it to my attention, I've never seen a bank or a credit union or any type of firm that takes your money and invests it and gives you back some of that for letting you use it. I've never seen it advertised that they pay continuously compounded interest. So why is it worth looking at? There's a couple of trains now. This would be the most that this money could make at this interest rate for this length of time. That's the most. So if you want to kind of compare what banks actually do pay you with what is the most that they could pay you for that money at that rate for that time, this is the most. So it's kind of like that upper limit. So it does have some value. Where were we with the other ones? 11.909. I bet you we're not a whole lot different there. Is it still 9.09? 9.12. Okay. So compounding quarterly, 11.903, compounding monthly, 11.909, and compounding continuously so it's not going to make any more money than this right here, 11.912. Questions on the use of that model or the development of that model from kind of the old standard compound interest formula? Okay, in the book let me read this statement and we'll translate this statement into a mathematical equation and then try to see what its equation looks like once we get out from under the differential equation. Newton's law of cooling, this is on page 529. So it's actually called Newton's law of cooling but it also, it's a change in temperature model. So it could be Newton's law of heating up but it also is Newton's law of cooling. So it works in both directions. Newton's law of cooling states that the rate of cooling, rate of change of temperature of an object is equals proportional to, there's a constant of proportionality involved, the difference in temperature between the object and its surroundings. Or normally it'll say between the object and the surrounding medium. So if capital T is the temperature of the object then according to this statement of Newton's law of cooling, the rate of cooling, the rate of change of the temperature, now we're gonna have a couple of T's in the problem, capital T is temp, lower case T is time, so that's the rate of change of temperature is equals directly proportional to, so there's some constant of proportionality. The difference to the temperature difference between the object, well the temperature of the object is capital T and the surroundings or the surrounding medium, I'm gonna call that M. So it might be the room temperature, it might be if you put something into some other environment where it's cooler than the room temperature or warmer than the room temperature, but that's whatever is the surrounding temperature. Differential equation, so it belongs in this particular portion of the book, slightly different, but also very similar to kind of the standard differential equation, which was the rate of change of Y with respect to T is directly proportional to Y itself, which kind of got us the standard exponential growth and decay model, so we expect it to be somewhat similar, but it's also gonna be slightly different. Alright, let's do the separable stuff. So what needs to be done to separate things, the capital T's and D capital T's from the lower case T's and the derivative of the lower case T's. Okay, and what else? Divide by T minus M, M is a number, so it's gonna be 70 degrees or 72 degrees or whatever. So the right side kind of looks like what happened when we were over here, but the left side's slightly different. What's next that we have things separated? Integrate both sides. Integral of 1 over capital T minus M, M is a number integrated with respect to T. It's getting a little redundant, right? I mean, hopefully that's something that doesn't require a lot of deep thought now, that we've done this so much. So we will be assuming in this case that the temperature of the object is larger than the surrounding medium, so it actually is a cooling problem. There's a possibility of a constant here. We'll throw that in on the right. Integral of K integrated with respect to T would be KT. We'll put the two constants together into a C. What's next? Okay, exponentiate both sides. E to the natural log of something. Is that something? E to the KT times C is going to be E to the KT times E to the C. E's a number, C's a number, so E to the C is a number. And add M to both sides. Very similar, isn't it, to this? Just kind of has that same look here. Obviously, B is not going to be the initial temperature, but it's going to be somehow related to the initial temperature and the surrounding medium. But if we just put the temperature of the surrounding medium in over here on the right, we've got the new mathematical model for Newton's law of cooling. So I think I have an example. Not there. So I was going to do an example because all the CSI shows are really big now, and every time you turn on the TV, it's CSI Miami, CSI in New York, CSI Fukeway. I mean, they've got all the big shows. But it's a little morbid, I think, to talk about when you're, you know, taking the temperature of the body when you arrive on the scene, taking the temperature of the body an hour later, you know the surrounding temperature of the room, and you're trying to determine the time of death. But that's a scientific way of applying this particular situation. I've got a little more benign example here. Scalding coffee, okay? McDivitt's Pie Shop's a national restaurant firm. I eat there all the time. Find that the temperature of its freshly brewed coffee is 130 degrees. The company fears that if customers spill hot coffee on themselves, lawsuits might result. Is that a far-fetched concept? Absolutely not. Many eating establishments have been sued and lost because they served coffee that was too hot and people drank coffee that was too hot. Go figure that. And then they sued the company because the coffee they were served was too hot and they won. So this is a realistic problem, which is one of the reasons why I picked it. Room temperature in the restaurants is generally 72 degrees. The temperature of the coffee cools to 120 after 4.3 minutes. I would think you'd maybe adjust your little coffee warmer so that it's not 130 when you're done. But anyway, that would ruin the problem. The company determines that it is safer to serve the coffee at a temperature of 105. How long does it take a cup of coffee to cool to 105? A little bit more fun than somebody losing their life and we're trying to figure out when they met their demise. Our model is... What do we know right away from this problem? We know the surrounding medium. 72 degrees. That'll be down the road. Eventually we want the temperature to be 105 and we'll know everything else except for T. We'll plug in 105 and we'll see how long it takes for it to cool to 105. So we've got kind of a point in time here. So at time 4.3, capital T is 120. That'll help the cause. What are the initial conditions in this problem? At time 0 when it just finished being brewed, 130. So at time 0, the temperature is 130. So there's our two data points. So let's plug in our initial conditions. Any questions on that from the wording of the problem? So capital T, we're going to plug in this particular data point first. Capital T is 130. That's our problem number. Kind of looks like we're going to have too many variables, but we get rid of them because T is 0. We don't know, okay, but it doesn't matter. So E to the 0 is 1, so we end up with 130. So B is 130 minus 72. Will it always be that in every cooling problem? Will it always be the initial temperature of the object minus the surrounding medium? Yes. Remember that, and that's what that coefficient B is. That's fine, or you can just plug in. It doesn't take a lot of time and effort to get B as 130 minus 72, which is what? 58. So our model is growing. As far as what we know, and we have fewer things that we don't know, what do you think is next based on how we've done problems like this? Find K by using the second data point. So that's provided for us. The temperature, 120. K, we don't know. T is 4.3, so now we're locked into minutes, right? Wasn't it minutes? Yes. So since we're putting minutes in for data, we've got to continue to put minutes in, or if we plug something else in, it's going to kick out minutes for us. So 120 minus 72. Divide both sides by 58. So that value is equal to E to the 4.3K. How do we get K out of the exponent position? Natural log of both sides. Natural log of E to a power. Is that power? Divide both sides by 4.3. So now we have our K value, and I think that we can then go to what we want to figure out from the question that's asked in the problem. What's K? 01911, same? No? Oh, maybe I'm wrong. 044401. Aren't those the same? 019 and 044? No. Commit is that. All right, so we have two people. Excuse me, two people getting that for K? Anybody else run that through and get that? Yeah, I did. So let's go back to the mathematical model, to the K, and we now know K. So the third data point is incomplete in the sense that we know the temperature, but we don't know the time. Any guesses before we start? Now don't think linearly, this is not a linear model. In other words, six minutes. 100, it took it 4.3 minutes to cool to 120. So it went down 10 degrees in that 4.3 minutes. It's not going to go down another 10 degrees in the next 4.3 minutes. That'd be linear, right? 22 minutes. 22 minutes? Yeah. So in fact, let's go back to the mathematical model itself. Doesn't the rate of cooling lessen because the object is now closer to the temperature of the surrounding medium, right? And it says in Newton's law of cooling, the rate of cooling is proportional to the difference in temperature. Now the difference in temperature is less, so the rate of cooling should be less. So it clearly is not linear, so we would expect it to cool at a slower rate because 120 is closer to the surrounding medium, which is 72. All right, so I've heard a 20-minute estimate. 22. 22, sorry. So we want the temperature to avoid lawsuits here to be 105. I think it's going to be less than that. And now our job is to solve for T. Take the natural log of both sides, 13 minutes. Pretty good. 12.81, right? I knew the answer ahead of time as I kind of cheated. I had seen the answer. I thought it was somewhere close to 13. So pour the coffee. This sounds kind of ridiculous, especially if you've ever worked at a restaurant. Pour the coffee and let it sit, right? For 13 minutes, roughly, before you hand it to the customer. They're probably not going to like that. So a better scenario would be to just kind of adjust the brewing temperature down. So maybe it's 110. And then by the time you pour it and put a top on it and hand it to the customer, it's at a safe level, 105. All right, but there is a pooling problem, and that's how it works. Did you notice some things in common about these problems? Didn't we have, like, initial conditions in almost every problem? And then we had some data point later. So we're kind of taking this curve. We don't really know what the curve looks like exactly, but it looks something like this. So we've got this initial data point and some data point later down the road. By the way, we do have in this cooling problem an asymptote. What would that be in the problem we just looked at? 72. Mathematically, let me set this scene up and then ask the question. So we pour this cup of coffee, it comes out, it's 130 degrees. We let it cool for 13 minutes, roughly, and it's down to 105 degrees. We leave it out there on the counter overnight. We come in the next morning. According to the mathematical model, is it ever actually 72 degrees? No. The first thing is asymptotic to 72 and it never physically reaches 72. Now, that's probably not realistic. If you leave it out there for a couple of weeks, it's not very appetizing, but it's probably going to be 72 degrees. So there is kind of some subtleties in these mathematical models that are actually slightly different from what we perceive to be reality, but they do model it pretty well. Let me see what else we needed to do today. Okay, I think I'm going to save that because we don't have time to do that. Anybody have a question, a web assigned question, or anything else that's pending? I said I wouldn't. We're not going to be able to get to a point where we can do much with this. We are going to start with the model. Let me draw the picture. And we're going to do a population problem. This is actually into the next section, but I think we need to start it to do justice to it. We do have a test win in here next Wednesday, right? So we'll have Monday to kind of wrap things up, Tuesday to review and talk about the test, and then the test will be Wednesday in here. So let's say that we have some limiting value or some maximum-supportable population, sometimes called a carrying capacity, but something is going to be growing somewhat exponentially initially for small values of T. But in the long run, because of finite boundaries of the city or county or town, water, supplies, all the roads and all that, we've got at some point in time some maximum-supportable population for that region. So as we get closer to that, the rate you can tell by the slopes of the tangent lines, it begins to flatten out. So as we approach this capital L, it begins to flatten out. This would be called logistic growth. And we'll start with, next class, we'll start with the differential equation that models it. There's quite a bit to getting out from under the differential equation. In fact, it involves decomposition into partial fractions. I can tell that that brings joy to your faces that with something we've learned, we actually now get to apply in this situation as well. So we'll start with the equation, battle the equation, try to get it out from under the differential equation, and then apply some logistic growth problems tomorrow.