 Welcome back lecture 26. We are actually kind of going to go off the syllabus a little bit today and get a little bit ahead but We have a test coming up. So I think it's probably a good thing that we end up just slightly ahead We've got some web assign questions on chapter 7 pending Let's get through kind of the the meat of 7.5 if we don't get to all the examples Today we don't have to because we have Monday to do this also And then we'll look at some web assigned questions time permitting. All right 7.5 talks about the logistic equation so logistic growth is very similar to Exponential growth or what I guess would be more appropriately called uninhibited Exponential growth early in the picture. So for short values short time runs or small values of T It everything looks pretty Exponential in nature But what we've got later on is some constraints in the problem that cause us to Have some limiting value or some carrying capacity and as we get closer and closer to that the population growth rate Begins to taper off and you can see that it's Flattening out to the point where it's practically zero as you work your way out to the right So we want the differential equation that illustrates that To show that as the population gets closer to L This limiting value that the rate of change of population Gets closer and closer to zero So I think you'll see that in the differential equation and then the task most of the task today is to take that differential equation and Solve it for p So the differential equation for the logistic growth kind of looks Very exponential Initially in fact if we stop there that is the model that gets us P equals P zero times e to the kt, but there's another term and I'm going to change a letter from your book This is actually in the book a lowercase k and they use another capital K down here I'm going to call that L for the limiting value Just to keep the K's and the L's Separate as opposed to a lowercase k in the capital K so the rate of change of population is Directly proportional to the population that certainly enters the picture But this extra term and we're going to look at not in depth like this But we're going to look at a couple of other models that When you look at the additional piece you kind of want to think about what that might mean to the rate of growth of population Well, let's go back to this fact right here if P approaches L What happens to this quantity in parenthesis? Okay, if P and L are practically the same that means numerator denominator the same this is one minus one which is zero So the rate of change of population gets very very close to zero So that's what that term is doing in this particular so it's allowing What we want to happen to actually happen in terms of the description of the population growth rate So it is a separable differential equation. It'll be the Testiest one that we've had yet. So it's going to take several minutes to get to a point where we can solve this for P So let me rewrite it so we've got room to do what we need to do so we would multiply both sides by dt that's pretty common and What divide both sides by? P and L's a number So this is some function of P as well So we want to divide by that also and you can do those both at the same time But I'm just kind of moving one term at a time So I think we've cleared everything out of the right side except for the K and the dt And all the terms that had P that were in a product setting on the right side now are in a quotient setting on the left side Anybody want to tackle this right now? At this stage P is the variable so it'd be like 1 over x dx so we can't do anything with the P value here or the P value here one thing we can do to avoid fractions within fractions is to multiply This fraction by L over L So that leaves us with an L in the numerator in this L in the denominator I'm going to multiply it by the second term to kind of clear the fractions So that'd be L minus P. Is that correct in the second one if we didn't have L's in P's If we had numbers in X's I guarantee you that most of you would know what to do So let's say for L. I'm just writing this off to the side Let's say L is a hundred and instead of P's we had X's and that was the problem that we were facing Prove me right here now. So what would you do with this if you were going to try to integrate it? What would you do? I'll give you a hint prior to integrating it Somebody said it partial fractions okay because we've got a linear term here and a linear term here Right, so we could decompose that into partial fractions if we do that. We'll get a natural log and another natural log Because they both have constant numerators. So that becomes the task at this point in time With this part of the problem. So Eventually we're going to integrate both sides. We're not going to do that until we decompose this into partial fractions L is a number The bottom and make you piece work Okay That's I like the thought because I I mean Those are things that you should say to yourself I don't want to do partial fractions because it takes too much work What if I distributed the P to the L and the P to the minus P Then we'd have a P squared down here. I'm all for it now. Now what? Okay, but okay, let's go so you would let you so all this is good thought stuff You'd let you equal the whole denominator or no Just just P squared. Okay. Well, that's going to be a problem because then you're going to have to accommodate the rest of the denominator somehow But As Katie said you're going to if you let you equal P squared, which is what we would have Then do you would be to P dp? We don't have a P in the numerator So that throws it out. The other thing is that we've got a term that has P in the denominator Which also kind of confounds that process, but those are good thoughts I mean to think is there a way I can avoid this those are those are valid things, but I think we're Missing some terms that we need in certain places and we have extra terms in certain places where we don't want them to be so I don't think that's going to get us anywhere But it's you know if it takes a few seconds to run that process through your mind, and then you throw it aside That's fine. Nothing at all wasted Believe me if we didn't have to do partial fractions on this I Would not propose that we do partial fractions So L's a number you talk about something that is going to look ugly. This is really going to look ugly Because everything's going to be a letter. So you kind of have to remind yourself that the L's are numbers So we're really dealing with something that looks like this Because now we're going to add some more letters How are we going to add letters? Hey, which is some unknown constant that we're hopefully going to find it actually works out pretty nicely A over P and That's a massive letters right a is a number B is a number L is a number. So the P's are the X's of old What's next in the partial fractions in this times P? So the denominators are now the same so let's equate the numerators BP that all-american gas station. What's BP stand for? British petroleum don't let them fool you. It's not American Just like I saw an interesting ad on TV It was about RBC Bank and then the email or the web presence was RBC USA Bank calm. Well, what's RBC stand for? Royal Bank of Canada. Okay, so don't let them fool you by never identifying what those things are They're not American British petroleum is not American RBC is not American. It's Royal Bank of Canada even though they say Add something USA at the end. I mean, it's not Not that I'm against them, but I mean they never tell you that So there probably is a reason why they never tell you that. All right. Let's distribute everything so What do we do from here Remember the P's are the Variables so we need to put those together, right? So if we factor out a P here Then that would leave what B minus a then we've got a a times L Well, let's equate things left side and right side. How many P's do we have on the left side of the equal sign? Zero we don't have any P's over here. So it must be true that B minus a Is equal to zero how many L's do we have on the left side of the equation? One so it must be true that a is One so that's easy if a is one and B minus a is zero then B is Also one so back to our Decomposition Which is right there? We can take this left side of the equation I told you this would take a couple of minutes more than a couple minutes to get there. You're not going to have to do this by the way, this is not a Testable test worthy type question takes too long, but I I have always tried in my classes and I will continue to try to do this that when we have a Starting point on a formula or function if we've had the background that covers how we get there to our end product I think it's worth class time to show how we get from starting point to the final equation Rather than just say oh, here's this mystery formula, you know trust me You know or trust James Stewart the author so let's if we can do it Which we can and I think it's healthy also to review things like partial fractions off and on You will see most of you in here would be my guess that you will take math 341 which is a full semester course in differential equations You will also use this quite frequently Decomposition into partial fractions in that course many of you will also take a second full course in differential equations Which is math 401 so you will see that use some in math 401 as well So it's not like we can learn this in 241 and it was good while we were in 241 But you know I don't want to clutter my mind with it later. You will see it and use it again All right, so for a we're going to put in one for B. We're going to put in one So the left side of the equation becomes one over P Plus one over L minus P Is that correct where we had L over P times the quantity L minus P. We've now decomposed that into Partial fractions So we should at this point be able to integrate both sides Isn't it easier to integrate sums and differences as opposed to products and quotients? The answer to that is yes, it is So what's the integral of one over P dp? I'm going to do something here that I haven't done and maybe some of you've had a question about this and It didn't seem quite right. I've always dropped the absolute value symbol because We're going to lose it eventually anyway, but let's keep it in this case To a point in the problem where we all agree. It's okay to get rid of it So I'm going to keep it and you'll see we won't have to on other examples Keep this because we're not ever Potentially going to have the natural log of zero or a negative number This one you might want to think through that a little bit so if this is this is the number L's the number. So if this is like a hundred minus X What would be the integral of one over a hundred minus X? Negative because we need a negative one right because we've got a minus X And if that's our you thing the derivative of you would have a negative in it So we'd have to put a negative there compensate with a negative out in front. So there's where that negative is coming from This is probably why I felt like it would be a good problem for us to keep the absolute value because it might seem a little more obvious that this could be negative especially if the Figure started with the population being larger than the limiting value it would settle down to the limiting value So let's keep it possibility of a C over here. We'll roll that to the right side The integral of K integrated with respect to T is KT We'll put a C over that This is typically where I fail to do something in this problem But I'm constantly reminding myself to do this in this problem. It works out a whole lot better We do have the difference of two natural logs What is the difference of two natural logs the same thing as? Right the natural log of their quotient right so If we have the natural log of A minus B We know that is the natural log of A minus the natural log of B, right? So if it's true in that direction and it's an equation it ought to be true in the other direction So if we have the difference of two natural logs, it ought to be the natural log of their quotient. I Don't want this one in the denominator. I want this one in the numerator eventually and you'll see why So let's multiply both sides by negative one It works if you don't do this, but it's not as Easy at the end So the negative term becomes positive and the positive term becomes negative over here That's kind of the reason for doing it and I know negative a constant is still a constant But let's just keep the negative C because we did start with a C and I guess there's a homework problem Or that's an issue too. Hopefully we'll get a chance to look at that So having started with C and we've now negated it So the natural log, excuse me the difference of these two natural logs is the natural log of their quotient Anybody see a reason why you might want the L minus P in the numerator as opposed to the denominator? See an advantage Couldn't we write this as L Over P minus P over P right because we have a sum of difference in the numerator You can't do that if you have a sum of difference in the denominator So that potentially is going to be helpful to us Once we have a natural log, what do we typically do at that point? Okay, exponentiate both sides So e to the natural log of that absolute value quantity is and over here we've got e to the negative kT times e to the negative C e to the e's a number negative C's a number so That becomes a number now. That's that's when we can get rid of the absolute value And in fact, let me just write it down here e To the negative C would be this if I want to get rid of the absolute value Can't I get rid of the absolute value by putting a plus or minus over here? Whichever one's appropriate right if it's negative. I wanted to gate it make it positive. It's positive. I want to leave it alone So we can now drop the absolute value because we're absorbing The fact that if it's negative we wanted to gate it positive we want to leave it alone So we can drop those at this point in time and all that whether it's plus or minus This is a number. This is a number all that can be absorbed into some Unknown number might be positive might be negative So we can carry them for a while. You don't have to carry them you can that that becomes kind of a Dead issue here at the end of the problem, but so let's do that division Which is the reason for negating both sides earlier in the problem. So L over P Minus P over P so that's L over P minus 1 Keep in mind the goal is to solve for P. I don't like where P is in this fraction at this point in time But we're going to be able to work that out real nicely Let's get at least the term that has P in it by itself So we'll add one to both sides So we now have the term that has P in it isolated Is it okay to flip the left side As long as if we've got two things that are equal if we flip the left side, we better Flip the right side. So the right side is really this over one, right? It's a sum so you can always put anything over one especially if we want to flip it So that becomes P over L when we take the reciprocal of the left side the reciprocal of the right side is that and Now multiply both sides by L So there's our final Mathematical model from the original logistic growth equation again not a test question But there are some important things reviewed there that I think are valuable plus we see that it's not some mystery formula So let's take a problem So that's logistic growth So if you see up here in the numerator you see a 500,000 then 500,000 must be the limiting population, okay? So that's pretty easily identifiable Let's take a problem that was actually in the book a little bit earlier This was in a prior this data set was in a problem earlier in chapter 7 But let's take this data and adapt it to Our purposes right here. This is the US population at 10 year intervals in millions and Let's pick some values that we're going to use as data points and then come up with what we think is a reasonable We're doing the problem. We can make up our own numbers So what do you think based on this data for the US population? Come up with a good limiting value for the population of the United States 500 million 600 million Population in 2000 was 275 million Anybody have a have a preference 500 million should be alright. We were determining this So we'll just do everything in millions so we don't have to actually write in 500 million and 275 million So the answer we get when we're done will be in millions So our equation So let's pick a couple of data points that we think will help us and let's predict the population Let's say in the year 2010 So what data points do you want to use and let's try to be a little bit discerning about? Dataset that we want to use to predict Here's what I'm talking about. Let's not use 1930 to 1940 To predict the population in the year 2010. Why not? I know this is a math class But let's put a little historical context on it. That's the depression. Okay, so Population growth rate during that interval Was probably not very large Therefore to use that as a predictor we might get the population in 2010 to be actually less than it is in the year 2000 so All right, so let's say we don't want to use 30 to 40. How about 1950 and 1960 I don't think we should use that data set either. Why not the baby boom in the 50s, right? All the servicemen got home from World War two. They were really happy to see their wives and Soon thereafter we had a population growth. I'm sitting here right now because I was one of those I Was actually an unplanned child My father told me I always kidded him He's passed away now, but I told him that I was the best mistake he ever made He didn't like to hear that but We probably don't want to use 50 and 60 because a similar reason 30 to 40 was a low growth rate 50 to 60 was a very high growth rate So I don't know want to use 80 to 90 90 to 2000 70 to 80 70 to 90, okay, let's do that. So here's our time zero We're starting the clock in our problem. We get to choose when we have a set like this So this becomes our initial population for our problem that we're starting and then we're going to go to 1990 which is time 20 and this will be our population later So our model so far So at time zero population is 203 million Time 20 the population is 250 million and We want to know the population at 2010 And that would be time 40 right 40 years from From our time zero. So we're used to plugging in data points. It's a little bit More cumbersome in this because of the where the variables are but just let your calculator do the ugly work So the population at time zero. Let's plug in t equals zero and the population at time zero So population is 203 So the nice thing about it. We don't know what k is but it doesn't matter because we're multiplying it by zero So it's gone. So this will help us solve for a E to the zero is one. So we just have a one plus a here So how do you want to solve this? Okay, basically cross multiply So 203 times one plus a and 500 times one Subtract 203. I'm going to write this out so we can compare it to something that's in the book Divide both sides by 203 Think about how we got here with the limiting value being 500 million Do you think in all cases that a? Is going to be the limiting value minus the initial population divided by the initial population Yes, it will be now That's something that if you chose to memorize that that could be useful and we avoid what we just did but it's not like it's real cumbersome So that's what a will be all the time. So What is that just? I don't know how much to use as a decimal here. What's How ugly is that decimal? Okay, let's let's use it 1.46 305 so that's our a value So we now have this much of the equation so we move to our second data point which is 20 years after our time zero The population is 250 million so we put in 20 for T We could cross multiply Let's go ahead and divide by 250 500 divided by 250 not very nicely is to kind of looked into that subtract one from both sides Divide both sides by one point and now that we've got the term that has the variable in it isolated We can take the natural log of both sides natural log of the left side is negative 20k Natural log of the right side we'll push buttons to get that Some point time and then we want to divide both sides By negative 20 so our second data first data point gave us a Pretty common a little bit more work in this case, but pretty common that it gives us that a or b or whatever P0 or y0 sometimes we just plug it in second data point usually gets you the k in the problem and then you Then the only thing you have left is t and you plug in new values of t to kick out new values of population So we have that k is What? 0.0 0.0 Agreement on that we're getting lots of those okay So now our model So k turned out to be a positive number, but in the equation we have the negative of k right So we can throw in a t value t is the only variable on the right side We plugged in L. We kind of estimated limiting value We found a and we found k So if we want t equals 20 Population is whatever then we plug it into this equation Population is what we're looking for And it's basically pushing buttons right it's already solved for p. We don't have to do any algebra Just push the buttons in the right order So this should give us the population on 20 doing 40 we want 40. Sorry if you did 20 by the way We better give what? 250 because we threw that in as a data point so that confirms that that's what I was really trying to do Just confirm the data not really so if we put 40 in like we're supposed to Then we should get the population since our time zero in the problem was 1970 what's it give us? 297 more Zero If you use the whole right so that's millions right so if we're just trying to get an estimate which We're using data to try to predict so I Think that's going to under predict because we're at 2009 and I think I heard last week on the radio Oh, I don't know why these things kind of catch my ear, but I think we're over 300 million already Zero like 316 Yeah, I don't know if it's that high, but I did hear 300 something so I know we're over 300 so it under predicted It's a prediction. Why did it under predict? Well, apparently the growth rate from 1970 to 1990 Is not keeping par with maybe what it was from 1980 To 2000 or maybe just in 1990 to 2000 the population growth rate is more so but it's a prediction All right questions on this All right, let's see if we can diffuse a couple of the Web assign Problems start with number three and I I think these are problems from the book I don't know exactly where they are in the book But what is three it's a differential equation? These easy to He brings to the t plus c Equal zero addition is bad. I don't like addition in a Differential equation, but in this case because we have zero on the right When you kind of add or subtract whatever to move terms to the other side It's nice because there's no other term there, so it becomes the only term on the other side So let's send this term to the other side Now sounded like in the discussion before class that some of you had had success on this and others had not So if you had success on this one What what did you do next? Well, let's talk about something before that. What what are we going to have to do? separate the Z's and the D Z's from the T's and the D T's so if you don't like the t plus z You can change e to that power to a Product right so we're stuck with a negative one out in front, but now we have an e to the t times an e to the z So if you had some issues with this problem, and you didn't go this route I think that makes things separable and there was some something else about a constant that they said Right, I think in the parenthesis. It says in the directions of the problem. It says Keep c negative in the answer, but we know that Plus the c and minus the c they're the same thing cuz c's unknown it could very well be positive very well be negative Do we need to keep going? Anybody want to keep going with this problem? Yes, okay. Tell me when it's to a point where you're feeling more comfortable than than this moment so we'll keep this here and keep this here and Divide by e to the z well if you divide by e to the z that's the same as multiplying by e to the negative z and There's possibility that you could negate both sides I don't know that that's going to be all that helpful, but it might be helpful because if you brought the negative over here Then you've got a negative that you need in the integrand, right? Is that those of you that had success on this problem? Is that something you did? Yes, okay multiply both sides by negative one Now we'll integrate both sides integral the right side is E to the t And we'll put our constant there which I think is again part of the critical part of the Solution is if we end up negating this then Leave it that way now. Let me see If they have a negative c in their answer, maybe they kept the negative over here You didn't multiply both sides by the negative Well, let's keep going with this because it certainly is valid to do and I think there's a good reason to do it What's the integral of e to the negative z? times negative one Integrated with respect to z e to the negative z and Back to the original problem. We had derivative of z With respect to t or in terms of t so it probably be Our what we want is z in terms of t right for our final answer solve for z in terms of t What's going to do that? Natural log of both sides So the natural log of e to the negative z is Negative z the natural log of this entire side It's a natural log of a sum which we can't really simplify right So if you chose to multiply through by a negative You would end up with that apparently that's not the path that the Whoever did the solution manual took So where were we right here? When we multiplied both sides by a negative Let's let's see that we can get another answer which obviously has got to be a hundred percent equivalent to this Chandler, do you have a question? In this well, we have to put absolute value for the natural log Trying to think if we could ever get a negative here with this I Think to be safe. We probably should put it because you've got e to the t Plus c C could be a negative number That could overtake e to the t So there's a possibility of a negative so probably To do that we should put I'd have to think about that some more but that that would to be safe. We probably should do that If we left stuff where it was Which I think is probably the case if they had a negative c What's the integral of the left side? Negative e to the negative z And the right side Negative e to the t plus c So this is probably where the negative c came about by negating everything Now we get a minus c And now take the natural log of both sides and negate both sides And again, probably the safest bet is to put absolute value there, but it sounds like it's accepting it without that so apparently this answer And this answer are Equivalent, but don't think about distribution. We can't do any distribution there with natural logs But the negative are the plus and minus the zane because we can do the natural log Because if c Let's say c is some arbitrary constant c could very well be negative 5 in this particular version Whereas down here in this one c would be 5 still it's an arbitrary constant and it ends up being subtracted here C is a negative number here c is a positive number So I don't remember the other one, but it looks like it's not going to happen today anyway So we still have a little bit more to do with 7.5 and when's the web assigned to Okay, I'll change that to Monday. So if you still have issues we can talk about that on Monday and We'll clear those up and review on Tuesday. Take a test on Wednesday Have a nice weekend