Logistic 相图行为与环境承载量

The did they say we couldn't be here? What? Ah, yes, I do remember. Ivy and and Ida are both asking asking to get the recording now. Oh, good. We can be more effective if you are the only student now. I'm so sorry. You really should that it's your time now. You're entitled to it. I just got into talking. I didn't watch the time here. You should shout, you should yell, you should holler, and say, "It's my time now. Class should start."Oh no, no, no! I didn't mean that. Okay, let's go forward. Now, finally, we'll be ready to actually take that logistic function and work out what would be the shape of the graph. So, could you actually recall for us and was there any loose end from the previous session? Because I actually took an excursion again and showed you how do we understand for complex numbers here. How do I take the in the natural log of even a complex.NumberYes, and what you remember is actually. Could you repeat for me? What do we have for the inside of the log? By the way, are you sure this is not what we're doing in the other class? Not in the calculus class. Yeah, I'm sure. It is in the other class. It's not in the calculus class, right? This wasn't the calculus class. I'm pretty sure. Oh, good, good. It's in the calculus class. Wonderful. So could you repeat for me? What exactly do we have on the inside?We're going to recapture the context. What we're actually doing is, what if you want to take you you want to find the antiderivative of something like that, one minus x, plus x squared. We did it in two ways. We can compute the square here, which is what I did, and which is totally correct. We change the denominator into actually the one minus, sorry, the x here. X minus half of.X that whole thing squared plus three over the four, and we recognize here this is if I call that the u here, it's very close to the arctangent, and we're actually recalling and what we did without using the complex numbers here.By a four third, so this is going to become a one plus four third, but I'm going to absorb that inside to give ourselves a two over the root three x and a minus the one over the root three. Sorry, squared here. Then, if I want to integrate this one, I'm going to force out this x here into a two over the root three x.But then if I do so, I've got to take away from here now what we have. It's only two over the root three left in the front, and then I can take on this because taking on constant won't mess up your differential, and this fits into the chain rule. So fundamentally, what you get is really two over the root three of the arc tangent. I want to do it because we need to really compare our results now. Of the two over the root three x minus one over the root. This is the entire antiderivative. If we start from here, makes sense. Yeah. Okay, but we did it some other way. We're recognizing the factor of the denominator. It's going to be actually the e i positive sixty degrees and the e i negative sixty degrees. I'm not going to repeat the process, but we did confirm that is the case in.So we factor out the denominator into this. Basically, we can do partial fraction now. It's the x minus e i sixty, and then it's not one on the top. Something, and that will get actually x minus e negative i sixty. And we're finding the numbers on the top in order to after we factorize it by plugging if I put at omega, I'll put the other the omega conjugate. So I plug the omega.We're getting is one over, uh, twice of a the omega minus omega conjugate, just twelve root three i. No, one one root three i, because twice of the half of root three i. On the other side, it's minus. That's what we're getting, isn't it? Yeah. Okay, beautiful. So we're finding their antiderivative, so which actually give us correctly. We're getting.One over the root three i on the outside, and I were getting natural log. Here you have the x minus the e i sixty degrees over the x minus e negative i sixty degrees here. And I actually made a big point: we can't add the absolute value anymore. Yeah, because we have to keep the imaginary part. So this is a you just put it into a parenthesis. That's all we do. Yeah, then we did the natural log. There was a.Real part and imaginary part. Very good indeed. So excellent, and let's actually expand out this complex number. This is where I actually go back to the graph. Now you have the x here, which is this vector in the real direction, and we're minusing this omega now. This is a basically when you get this vector now. This is one minus. The sorry, that's the x minus. This is the origin of the universe.Negative to the x direction, and negative to the x zero. So this complex number is x minus zero ei sixty degrees. And now likewise, you would have a sorry that one. And that is, this is going to be our sixty degrees, sixty degrees. Now that's the x minus the e negative i sixty degrees, correct.At this moment, here we're realizing if you're dividing them now, you're getting actually the twice of that angle. And so, well, at least we're noticing this ratio has a magnitude of the one—that's for the magnitude—and has the angle. If I put at the theta, we're just getting negative two theta i because this is a negative theta, and you're minusing another theta.Yeah, so when we take the natural log of this quantity, in fact, all we're getting is simply going to be a negative two theta i over a root three i. There's nothing else. That's the only part that comes out. Makes sense. I'm still a bit confused about the denominator. The denominator, you mean the root three i? Yeah. Oh wait, never mind. It's coming from the outside. It's coming from here. So is everything crystal clear? Yeah.However, I'm in trouble because somehow we gotta. It's good now because this is really going to be the negative two over two three. The imaginary part cancels out. We're back to the real, and the theta is simply the arc tangent of something. And we're working out what is that arc tangent. We do realize though, and that numerator it is simply going to be the root three over two. And this is arc tangent of the root three over two.Then divided by this quantity here, which is our x, that minus the one half, isn't it? Can you repeat that? We're working out from the geometry here. What is that tangent theta? Wouldn't that tangent theta equal to the rise over the run?And remember this is a unit complex number with a sixty degrees over here, so don't we know that the vertical component will be the root three over two, and the horizontal component here is one half, right? Yeah. So remember the whole point from the origin all the way to the point P, for example, I call that P, it's actually x in length. So don't we have the rise over run simply written in such form? Yeah. So can I.Write a script concerning Chinese我 theta 现在作为这个量的二分之一。我高兴，因为二分之一现在出来了。我的问题是，现在它并不完全像那样。它们并不完全相同，但它们是的。实际上并没有什么问题。但然后我们需要理解为什么它们实际上是相同的量。嗯，部分一，为什么我们试图吸收那个负的？因为这显然是。Not a negative angle. Well, the earlier is a positive angle, right? But now we seem to be getting a negative angle. But remember, though, when you find the anti-derivative here, there's an arbitrary constant attached to it, isn't there? So in fact, it's adding arbitrary constant C now. We're also adding arbitrary constant C, although these are not the same C's. So I'll be smart. I'm going to actually pull out this two over the root three, and I'm giving an arbitrary constant which is pi over two.And then I copy this minus arctangent of the root three over two divided by x minus one half. I'm allowed to do that because I'm free to pick an arbitrary constant. I just happen to pick that one. Then from the graph, or really just fundamentally anywhere, if you have a theta equal to.Our tangent of a complicated number here. We're gonna call that u now, and what is going to be the one half minus theta. Can you also write this angle here as our tangent of something? I'm really asking you now. Now, if you know the tangent of the theta, what do you know about the tangent of its complementary angle, half pi minus theta? ```You could probably use a trig identity to find it. Indeed, in fact, a easy graph here would really illustrate it very well. And why don't we pretend that it's a cube because it's easy now, and the tangent is u to one. That's the easiest way to construct the angle where it satisfies its arctangent u, right? Yeah. And this angle is complementary to it.That's just half of pi minus theta, isn't it? Yeah. And then what is the tangent of this angle, the complementary angle?Can you see that from the graph? It is still the opposite over the adjacent. Oh. Or you can tilt your head and think about this as your horizontal axis. Now it would be the rise over run.And this would be the so-called rise now, and that would be the so-called run. So it would be one over u. Indeed, so in fact, you can write it. By the way, I'm going to highlight it. It's a generic fact here, meaning if you know that the angle, the tangent of the angle now, and you know the complementary angle's tangent is just the reciprocal thereof. So this is our tangent of the one over the u, isn't it? So can I write it out?This is indeed a complementary angle, so that equal to the two over the three, our tangent. Of the reciprocal of whatever that inside, which bumps it up to the s minus the one half divided by the root three over two. Correct. Yeah. Now compare very carefully. Is this the same as that one now? Are they the.Yeah, they're perfectly the same. You might wonder why do we even go through such a securities route just to why can't we just do it that way? You can certainly do it that way, but I really want to pave the way for our future selves here. When you really learn higher mathematics, you really want to be not afraid of complex numbers and their logarithms here.I also really want to bear out the important concept: as a complex number, you do not add the absolute value here. If we do, we're actually losing the imaginary part. Yeah, you said that last lesson too. Ah, beautiful. Yes, I'm I'm repeating it. That's my principle. If something is important, I say it three times. Very good. So this is, and we are finally ready to go back to the logistic function now. Fortunately, logistic function is far from being. So complicated. We solve that all the numbers are real, that denominator is fully factorizable, and let me quickly recap. Here's a logistic function, which we start off. One second. We start off by actually doing the dp over the dt, and we got that differential equation that's a negative of, or I wrote it as positive here p, and the carrying capacity minus the p, and we separate out into the dp.And that's the p, and times the carrying capacity minus the p, and in fact, I'm going to artificially multiply both sides of the equation by the carrying capacity, which is equal to the alpha times the carrying capacity. And then times the dt, and we integrate both sides. We break it down into partial fraction, which is really the dp, and inside you got actually one over the p, and plus one over.The PC minus the P now. I'm scaling on both sides of the equation by the PC, just so that we don't end up with a fraction here. What is equal to alpha PC. I'm going to actually refer to alpha PC here as a new a here. It's just a new parameter. It's a real number, given by the system. We don't need to solve it. And that just equals the AT and plus a certain object constant C. I'm talking quickly because.We have let's just review. We have gotten here now. So the the equation you end up the natural log of the p over the p c minus p. But here we do need to add the absolute value because all the numbers are real. And that actually equal to the e of the sorry that's not there, and just the a t plus c, correct. And at this moment, we want to exponentiate both sides of the function.Get rid of the nasty natural log. So what we're getting is actually e of the right hand side, a e plus c, or or the adding the exponent really amounting to multiply the exponentials. So this is going to be some kind of a constant here, which I'll call that the k to begin with, would actually equal to that p over the p c minus the p. This is a generic solution, and we have an arbitrary constant k.Right now, the k is an arbitrary number. It's a real number. It differs from the a and the uh, the other NPC and the other parameters. But the k is not defined by the system. Yet, unless we introduce initial conditions. Do you follow? Yeah. Do you see any issues? May I make that move?There is a bit of intentional bullshit in this. I can explain it both sides, all right. There's no issue there, and in fact, I also know the k must be greater than zero because e to r c. I name that as.Okay, now which is always greater than zero because in itself it's already exponential. But is there any issue? Tell me what you're thinking. It looks fine. I.I can't really see an issue yet. Oh, it's a it's a tiny detail. How did we throw away the absolute value? When I cancel the inverse pair between the exponential and the logarithm, that's all right. But where did I leave the absolute value? Must we still keep it? Oh, yeah. Oh, yes, indeed, we still need to keep it because we're dealing with the real numbers here.So if we're crystal clear about keeping the absolute value now, and let's think about how do we take off the absolute value and expose the p. Eventually, we have to isolate the p now. And I'll let you go ahead because you really need to discuss the situation. Meaning, we have to actually look at the case where the p is less than zero, where the p is between, is greater than zero, but less than the p c minus a p.It is greater or equal to. Oh no, that's not defined yet. So for algorithm, or p is strictly greater than the PC. These are actually three scenarios: part one, part two, part three. Well, as the real model for this population now, we know part one is entirely not not legit. We can't have negative population. Part three is unstable, but it doesn't mean we can't. We can't have a instantaneously a population greater than the current capacity.It is just going to be that populations will kill itself. Yeah. But good mathematically, though, I still want you to write out what this equation, star equation, becomes in one of the in each of the scenarios respectively. And isolate the p for us for each of the cases. AndKeep me posted. Once you have some partial results, I'll put it on the board for you. So, so I see what the equation becomes for each scenario. Yep, with the goal of isolating the P and actually eventually solve what is the population dependent on time.`````` 我。中文嗯。What do you mean by solving the equation for p? Do we put p to one side of the equation? Yes, we isolate.It means we what we always mean by isolating the p.``````Because of the absolute value, there would have to be like two equations to solve for because one would be positive and one would be negative. Right, but to begin with, here within the three different ranges here, if the p is less than zero, the first job is to take off the absolute value. So how does that?Come out, when I throw away the absolute value now. And if the p is less than zero, inside is a number positive or negative. If p is less than zero, then it would be negative. Correct. Now how would you write it after it comes out the absolute value? Along here, Elaine, if you're taking the absolute value of the negative seven, how does that come out? Seven. And if I know the a equal to negative seven, and you take the absolute value of the a now, how does that come out? Seven. I mean, you write it in terms of the a because it could be some other negative number. Negative a. Yeah.That's right. So for negative number, when you take the absolute value, you have to take on if it comes out of the absolute value sign. So this really becomes a negative p, and then the p c minus a p, right? Yeah. And then you can go ahead and solve the p now. You can isolate it. Tell me what you'll get him.Do all your calculations very carefully. Don't trust what I put on the board because I may be bullshiting you.```I got the same thing. Alright, we.We can beautify that a tiny little bit, and it's not really important, but nevertheless here, and I'm hoping that it might actually give us a better sense here by dividing both the top and the bottom by this k e a t. So it's going to actually become the PC that's the carrying capacity, and on the bottom you just have actually one minus. Could you move a good and then your minus one over the k of the e of the negative a t.For that purpose, here, why don't we actually call that one over the k? Because if I have done so, all of these are one over the k now, and eventually this will become k. Just to make it easy, because the k is to be determined, we don't even know what it is now, so I can feel free to assign anything. Okay.We're still good. Yeah. Alright. So knowing this is our final answer now, we really want to make sense of it. Eventually, we're going to actually graph all the three solutions from each component and put it on paper. I know this year I've trained that into a one over k now. Okay, keep that in mind. I well, let's actually see what's going to happen under scenario number two. By the way, this is the usual meaningful. Part of the solution. So I'm gonna actually write it in purple, and you go ahead first, and we repeat the same procedure. But step number one, if p is within the range of, it's positive and yet smaller than the current. Sorry, I didn't mean that. I mean that it's smaller than the current capacity. I do not mean smaller than half of the current capacity. But anyway, and how does this inside come out of the absolute value? It would always be positive.Meaning. We wouldn't need the absolute value. Good, good. Meaning, we'll simply copy it like that. Would it equal to the one of the k of the e eight t now? In fact, knowing eventually, we probably want that a whatever the number on the top here. Then I would suggest we're going to directly write it in this form, judging from what we had done earlier, reducing it to a negative eight.So I would immediately put everything on the denominator, and then go ahead again, isolate your P, and see what you're getting.`````````中文`````` It's yours negative. In web. Ah, I got the same thing, but mine's the negative version.Then you have to do that more carefully. Meaning, which term we're keeping on which side of the equation.中文```Yeah, I got the same thing. Good. Alright. Well then, later we shall plot all of this, and apparently.They are not in the same form, and in fact, they behave drastically differently. You see, because one doesn't have a vertical asymptote, the other one does. Let's finally solve for the third one. Well, apparently, we notice here coming out of that one, we're getting exactly the same because right now it's also negative, and that's actually one over the k. And I'll put that directly on the denominator. So at this moment, the equation is identical to that of scenario number one. So we realize we got two alternatives.Solutions here. This is combined version one and version three, and the solution lies within their separate regions here, and that's the version two. Do we agree? Alright, now we're ready to actually get a visual plotted on the graph. Now, don't do best modes yet because this is decently easy to graph. We're gonna actually do it by hand here. So, want to actually get do a better job, giving us a decent coordinate system.Hey, come on. Shoot, shoot. What? Alright, so this is our coding system.And that's the T, and that's the P here. And we're going to separate out into three regions now. This is our carrying capacity, and that's a zero. And I'm going to actually separate them to begin with. The solution don't lie within the same region, right? And that's going to be sort of a separation. Above that's region one. This is region two. And that's region three here. Let's grab the well behaving one of the region two first. When we plug in the t equal to zero, what where do we start? That actually depends on the k, but we know for sure the k is greater than zero. So in fact, if we plug in the t equal to zero, the p at the zero point here equal to the pc over the one plus k. Oh, that's very meaningful because it gives us a compass to find out.How to locate? How to select the K? If in the future we do know what is the beginning population, we're not reaching the carrying capacity yet. We start from somewhere in between. So, for example, if this is my PC. For example, if I actually tell you that the P zero actually when we starts at seventy percent of the carrying capacity. If that's the case, can you find out what does that mean to the K? What would be our K?k equals three over seven。嗯，brilliant，that's very good. k would end up with three seven.This is a scenario we're plotting, and can we notice here? We're going to actually draw that entire graph now. Using this concrete example, the k equal to three sevenths now, and we do not have to actually find out whether t is increasing, whether the p is increasing or decreasing, what happens at infinity, etc. So let's see if the t is greater than zero, and as t increases here, what's going to happen to the.What happens when t is approaching positive infinity? Maybe we're going to find out the asymptotic behavior first, because it will give us maybe an asymptote to confine our curve. If there is any asymptote, what happens when t is approaching infinity? It approaches PC, indeed, because if the t is approaching positive infinity.This negative exponential would be approaching zero, the denominator would be approaching one, regardless what the k is, and that just means my curve would be gradually approach that PC. And indeed, that's functioning like a like what is sure that it's a carrying capacity—it's a maximum population that's livable within the pound. Then what's going to happen though when the t is less than zero but the t is decreasing? What to begin with? WhatHappens when the t is approaching negative infinity. It approaches zero. How do we know that? Because this exponential part will be approaching. Paul the infinity, right? So that means our denominator would be approaching infinity, and but it's a denominator when you divide the PC by it now, indeed it would actually approach zero. In order to approach it, we have to bend it the other way. That's why I just intuitively change the concavity of the curve. We flip it from a can't give down to can't give up in the negative region, but I want to be more.Accurate now, so can we find out the curvature of the curve here? Meaning, and where is that concave up? Where is that concave down? Meaning, we want to find the inflection point. Can we use the derivative? Absolutely, in fact, we have to use the derivative to know whether the derivative is increasing or decreasing. I don't suggest you take the second derivative; it's not necessary because you can see it's pretty complicated, right? It's nasty. I don't even suggest.You take the derivative from here. Why do I not suggest that? In order to decide the concavity, we do need information from the derivative. But I said, I don't need to take the derivative. Why not? You think about where the whole story comes from.``````What are you thinking? You remember where did we even get that differential equation? What is the beginning point of the model? We found a differential equation for the change. That differential equation is what.Start from here, right? Sorry, this is that differential equation, isn't it? Yeah. Doesn't that already give us what's a derivative? Yeah. Does that automatically tell us the concavity? Yeah. How tell us? Meaning, within what condition? Oh, under what condition?That indicated the population is coming down. It's increasing or decreasing, but it's coming down now. And what condition would be coming up? Knowing the derivatives is that, well, there's only one issue though that derivative is not written as a function of t, so you can't say, "Oh, that derivative is a how did we define coming up." That means the derivative is increasing. Yeah. And coming down is defined as the derivative decreasing. But the implied behind that would be it's increasing with time and it's decreasing with time. We don't have that direct information because the initial differential equation. I'm gonna copy it down here. Is that actually the p dot here actually equals to some constant here of the p times the pc minus the p. It's written in terms of p.It's not written in terms of a t, but I declare it's easy to really just reason through even that differential equation written in terms of the p. That's fine. We could still easily go from there to get whether the p derivative of p is increasing or the equation was time. I won't say it. I need you to reason this out.```Tell me what you're thinking. Don't we know P? We know P. You can just take the derivative. Like you said, I don't want to. We don't need.To really take the time derivative of that quantity, we actually not only know p, we stop that from. We also know p dot. We already know what is a p derivative. I just need you to translate it into where that inflection point is, where that S curve here. The roughly speaking, got actually logistic curve. By the way, this is got logistic curve now. It's an S shape, and I want to know where that turning point. No, it's not the turning point. It's the turning point of the derivative. It will be where the.Inhabitate changes. They do ask you this kind of question on the AP. So I need my students to know.```Hey guys. You forgot we sent out the message two weeks ago that the meeting time is eight forty now. That's a regular shift. Oh, I see. Thanks. Okay, actually for today, could you come back at eight forty five? Sorry, because I'm a little late in our in my previous sessions. Okay, so thank you. See you guys in fifteen minutes. You like talk to me.We can find like when the derivative turns negative. No, we don't. The derivative is given now. Can you just follow my lead? You know, this is a derivative. It's written there. You just need to make sure you know whether that derivative is increasing or decreasing regarding time. Once you know how, it takes only one second to realize what's going on. Elaine, we grappeded earlier. You asked me for what's justifying this model.When I raise my voice, I'm trying to be dramatic. I'm not trying to scold you. That's not my intention. And what you're doing super well. So did we not start from understanding the behavior of the Pdot intuitively? We started by I was justifying if the rate of change of the population increase can't be very fast either end, whether the too much of the population, which is the PC, and whether it's zero. So it's actually coming back to zero, but it would actually peak in the middle. So to.Oh, yeah, I remember that graph. Right, very good. Except for the horizontal axis, is not time. The hor horizontal axis is actually population itself. Nevertheless, it's very informative. Whatever this happens here at the inflection point, we're reaching because that just means the population at the rate of the population increases. It's increasing before that, it becomes decreasing. What do you want to p dot, correct? So obviously, at the inflection point, we're reaching.Human maximum of the P dot, aren't we? Yeah, and when does that happen? Combine the two graphs here. This is the P dot as opposed to P. We found the maximum before, and it was PC over two. Indeed, that comes with a parabola. Very good. So the maximum necessarily happened the PC over two. Does it hold here?Does that apply? Does it tell us for sure the halfway between the zero? I don't mean halfway in terms of the T. I mean halfway in terms of the population size here. Exactly at the halfway of the maximum population carrying capacity. Now we're having the inflection point. Does that make sense? Yeah.So we know a lot about the S shape now, but then can you prove to me this S shape is entirely symmetrical about that half of the carrying capacity point? It happens at the some time t, which I'll call at the t one half, and the region of population one half of the carrying capacity. And I want you to prove rigorously this whole S shape is entirely symmetrical about that point. Meaning, if you rotate rotate the handle by one eighty degrees, now we're getting the curve.Itself. I wanted to make a rigorous argument. Because if the curve equal in shape now, it means their derivatives must be also totally the same. Meaning, if I'm going to actually find this point, which is actually this much and the distance away from the semi-population or halfway population, and then you could find another quantity here, which is also exactly sharing the same slope. And once again, you don't really need to take that derivative here. I need you to justify from this this graph alone. And once again, I'm going to remind you, this is now the function of t, and that horizontal axis is t.`````````Again, show me how you think about it. I'm thinking about maybe using the derivative because then at two points the slope would be the same. Indeed, at what kind of two points the slopes are same from this graph here? Well, then obviously, if you want to hold the slope the same here, there are two populations that you're going to guarantee the same p dot.And these two populations from the nature of the parabola, can you make an observation? What kind of relationship? Ah, PC over two minus x and PC over two plus x. Indeed, they must be symmetrical. That's the nature of the parabola. It doesn't really come from analyzing our solution. Comes from going back all the way home to the differential equation which gave rise to this logistic model to begin with. That means these two points here must be symmetrical about how.Halfway population. And yes, indeed, they average at halfway population, which actually means mapped on back onto that graph here. So if you do have a mapped onto the halfway population now, you got to actually a population here and a population here, and they are symmetric about halfway population. Then necessarily, curve is experiencing the same instantaneous slope here.Yeah. And if that's the case, we have proven indeed this whole S curve here. It's entirely symmetrical about the point of symmetry in the middle, halfway carrying capacity. Crystal clear. Yeah. Okay. So these are the fundamental behavior and characteristics of our solution to this logistic growth, and it all makes sense because basically we're saying. You start from somewhere, and we're going to always be depending to that carrying capacity. Nature just tries to proliferate itself, and if that's the case, it's bounded by the carrying capacity simply because of the exhaustion of the environment. Now my question would be: What if I start with a population that's even higher? For example, a peasant just dumped a lot of a fish—the fishlings into basically very, very baby fishes.Aren't into the pond now? And if that's the case, I mean by fishes, I mean many kinds of many families of many species of a fish fish. But if I do so, and that what's going to happen, that that that must mean if I start from the region one now, I need to go for this solution, and that's a different kind of a curve. But we still ask ourselves such such questions now, where what happens when the t is greater than zero, what happens when t increasing, when t decreasing.As importantly, what's happening? But more importantly, I also want to know: is there any vertical asymptote to begin with? It's a rational function. Yeah, it would be one. Oh, yes, indeed. Can we identify? I'm going to use a different color. Can we identify what is the vertical asymptote to begin with? Hey, this one has a come on. It has actually a vertical asymptote.Where is that located?It would happen when k equals negative eight t equals one. Absolutely. Skyworker and Miguel, would you be kind enough to wait for me for five minutes? I'm sorry, my classes are late today.I will be back to you to your group in five minutes. Thank you, thank you. So, and you're right. So we identify the location of the vertical asymptote. It's really the one. Basically, you can solve for this now e of the negative k p just equal to one over the k, or you can reciprocalize both sides now. It just means the e of the a t equal to the k here. Meaning your t it just could be the one over the a, n times the natural log of the k. Whatever it's well defined in.Because we know for sure k is positive, so it's always meaningful to take the natural log. We can find such a place, whether it's greater than zero or less than zero. That depends on my k is greater than one or less than one. In our particular scenario, we found the k is negative, right? I mean, not the k is negative, but the natural log of the k is negative because the k is less than one. Yeah. For that purpose, here we end up with actually a positive, ah, sorry, a negative.That means my vertical asymptote somewhere last year. Now once we know the vertical asymptote, now we can look at what happens when t is greater than zero, what happens when t is increasing, and what happens when t approaches infinity. And what happens when t is. It would be PC over one minus k. Yeah, which actually gives us the one minus k gives us four fourth, and when you reciprocate it, you get seven fourth. So indeed, we start off from a number beyond the current capacity. That makes sense. And then decreases for a little. Well, the lah, I'm not saying that. So what happens when t increases?It would approach PC. Indeed, it decreases and eventually approaches PC, and it's going to actually be approaching the vertical asymptote in the part of the direction. Now, what's going to happen when a t crosses that critical point?This is, I'll call that actually the t critical, which actually equal I'll call that t v, meaning it's the t that gives us a vertical asymptote. Remember that equal to natural log of k over the a here. What's going to happen when you cross that line?Usually, the decision is between whether we flip the side or do we not flip the side. Meaning, that curvature is starting also from the positive infinity or does it flip the side and start from the negative infinity now? Which one will be the case?And tell me how you're reasoning this out.Evan, can I leave this as homework? I want to completely graph the function, and after you do it by hand, now be sure not to cheat because you need the skill, and then you can check with decimals. What's interesting is that, in fact, the curve, the equation that combines with region one, the three, it also combines the curve together.Don't differentiate. Naturally, when you finish that the curve here, it already spans between the region one and region three now. I'm sorry. Yeah. Okay, and understand though why that actually that's meaning for for the real model within the fish pond. And keep up a good work. Really. Bye. Okay. Bye.

Logistic 相图行为与环境承载量

Logistic 相图行为与环境承载量

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