Logistic 模型解与图像行为

It makes it easier. Eddie, good to see you back and happy new year. And your dad gave me a very good report. I'm genuinely impressed. And now I begin with usual. Did I leave any loose ends unfinished from the previous session? And other any questions or insights, maybe even. I do remember in the previous session I left a task that we're going to actually begin with.Whether somebody ret teaching me back the material, the logic of the last knowledge point we covered. But I'm gonna actually defer to your questions first, because that may actually help you do your presentation. For the slope field drawing, it was symmetric. But I was confused because I thought there was only supposed to be one side.What do you mean? You have to actually identify. These are nonsensical statements taken in general. You have to narrow down to what kind of function. Because not every single slope field is symmetrical, though. When we made the slope field of y prime equals y squared sine x, the graph was symmetrical about the.Prime equals to, say it again, cut le. Um, y x squared sine x. Y x squared sine x. Yes, we understand the derivatives of the function regarding x and regarding y. Therefore, the original function is an even function regarding both the x-axis and the y-axis. That actually means your graph must be actually, and I think we actually plotted it. It works like that, right.Yeah. Well, then you're asking. Like, which side would we keep? Oh, that's a beautiful question. Well, why do we have to among among how many sides are you asking? Four or two. Two, like this one and this one. Yeah, exactly. Well, in fact, all of these, and that's not the only two curves; they're infinitely many because you could always.Didn't we end up with actually infinite many of these? There's arbitrary constant there. Yeah. Right. Well, then you're really not asking, ah, between the two, which one do we pick, or but rather, be among the infinite many, which one do we pick. Right. Yeah. Yeah. So this is the first order. So we're actually having only one free parameters to determine which one we pick. Now we actually use the initial condition. For example, this is.It's eyeballable. I can really say, choose that x equals to, hmm. Oh, I can't say that. I can find a point on the curve, for example, x equals to zero, and y equals to one. For example, I nailed the this initial condition now. I need I nailed the y intercept, and the in the solution we came up with.Can we find out which which is specific curve do we pick now? Yeah, because immediately it nails it to one single curve, and then you must have a this curve now. But then, do you still have the question between the top and down, which one do we pick? I don't think the question exists anymore because the bottom curve here doesn't even go through that point zero one, meaning it doesn't satisfy that initial condition. Oh, does that really make sense? But if we're just adding a constant, why would the bottom one show up? What do you mean? The bottom one doesn't show up in your solution. There are infinitely many. I just don't remember what they looked. Maybe they're going to look like this. There are infinitely many of these. They're not two. They show up because they satisfy that different.The differential equation. There are infinitely many curves satisfying that differential equation, unless we know which specific point that curve happened to go through. Oh, oh, oh, I get it. Do you really? Yeah, they both are here because they fulfill the equation. But with this condition, we can narrow it down to one. That will be the solution. That's exactly right. That's exactly right. Ivy and Eddie, are you with with us?So like the, so it has to match the y-intercept. Whether the given condition, you're right. Then narrows down to only one. But I have another question now. We just solved the logistic function, right? And we actually I give you the initial condition. For example, this is a p prime as a function of time would equal to the carrying capacity, da da da da da. ButAt the moment here, I'll just say this p, and apparently the current capacity will be five hundred minus p. That coefficient, fortunately, is just going to be a simple one. It's a two. Quickly, guys, why don't you solve that differential equation for me? You will be asked repeatedly to do this, perform this every year on AP. So go ahead. We practice doing this for quite a lot of time now. I recommend you don't even need to do separation of the variable. Da da da.You just remember what is a form of that differential equation solution now. What is the P S function of t, and if you still remember roughly what is a form, then we can actually apply some initial condition to figure out the special coefficients. But if you're just solving this, there's no way you can know. Then these number a numbers must really just stay as unknown constant, and then we're going to investigate and figure.It out using concrete numbers. Um, is it a polynomial? Logistic. This is a logistic population growth model, isn't it? Oh, I used it on the p prime. It's a polynomial regarding p. Yeah, but that doesn't give us a slope, does it?I mean, sorry, it doesn't give us a slope field, does it? Actually, it does. Can we draw the slope field? But however, though I actually want you to solve it first, just so that you have something to compare algebraically—not graphically, but just algebraically—we learn to do this.The one I'm giving you—it's a little quirky, but it is at least easily inseparable. 嗯，你看，Blood the P variables on one side, and the T variables on the other side. Yeah, I cried. Agree. You're welcome to write it on.这个是总评，这个是总评，我们刚刚已经覆盖了，然后ivy是实际上在做最好的事情，她已经离开，然后她已经配合了。Over two p times five hundred minus p, and then equals to d t. Indeed, I recommend that we keep the more complications on the simple side. It's easier, right? Definitely.嗯。嗯，One side would be two T. Yes, absolutely. And there's.This object comes in there, which I'll call that the C first.Uh, do we need to do the partial fraction? Indeed, very good. Uh, so we could separate to one over p plus one over five hundred. Excellent.Very good.嗯，嗯，呃，我们可以把 p 接近于零。 Oh, you actually missed a little detail. When you do the partial fraction, we'll just try Adam. Find the common denominator. Are we getting what we're supposed to be getting?We would be getting five hundred, right? Yes, that's right. If we don't want to any complication, we can pretend that there is a five hundred. That means we need to scale up the other side of the equation, correspondingly. So this one gets to be ten thousand. Sorry, one thousand of the t plus a c. It's not the same c anymore, but c is a symbol.So, some kind of constant. Oh, sorry, we haven't done the antiderivative yet. So Ivy, watch out for that little detail. So you should just scale it by, oh, five hundred. Right. To match with the five hundred on the numerator on the left.Then you add the two partial fractions. Then what's the antiderivative?Natural log of p minus natural log of five hundred minus p. Very good.Antiderivative of the two partial fractions is natural log of the p minus. We're gonna directly use identity, doing the natural log of the p over the oh come on, hey. Over the five hundred minus p, brilliantly done. And at this moment here, there's a little tricky detail. What else should I write on the left? And I told you to watch out not to.Right, if we're operating in the domain of complex numbers. But you don't do that on AP, okay? AP does not expect students to know anything about complex numbers. It's our own trick, but here we have to do it right. Is the antiderivative of one over something just a natural log? You have to put in the absolute. Indeed, so we can actually combine the absolute values into.To a single absolute value. So this is natural log of that one equal to one thousand t plus the c now. Then finally, how do we resolve it? But before we go on, though, I'm going to give you the initial condition. Initial condition would be it actually go through the zero and seven hundred. Ah, yes, and we realize it's already above the carrying capacity. So we're anticipating. I'll let you guys finish this. Remember. We discuss this in great detail. There are three regions. There's a region between the zero and carrying capacity, which is recognizably five hundred. We know we're starting higher than that. We're starting roughly here, and does anybody remember what would be the shape of the graph?```Are you seeing the graph like above the carrying capacity, or just the original one?The initial condition, knowing when t equals zero and your p is equal to seven hundred.Do we like plug in zero and seven hundred?I'm actually assuming. Ah, very good. Go ahead. I'm gonna give you plenty of space. Graph it. Put it in. Just for like the Asian, so like, um, I think it was like something like this. But yeah, it's yeah indeed. It's an ad shape. If the initial condition is two hundred, for example, according to your graph.And that as shape, yes, excellent. It like balances out around like the. However, Eddie, you haven't drawn that very accurately. So first of all, where is that point? Eventually. I'm sorry. Ah, could you repeat what you just said? Yeah, I'm asking, and how did you do.Determine that point. Does that point just stop there? Not very likely because according to your graph, it's probably gonna continue into. That's mine. I didn't write that. Meaning you didn't graph this. I graphed it. Oh, okay. Oh, then you remember it a little off. May I erase it? Elaine, Elaine graphed like the bottom part. I graphed like the top part. Okay. Good. I will finish it. I will rather just continue.```Eddie, are you putting in the graph here for the lower and upper regions?Anybody, if you actually remember what happened, or if we don't, let's resolve it. I remember there was like a line. That is the inflection point, where the concavity changes. Good, very good indeed. To find that one, we can look at the derivative, right? Hold on.And don't break off from the line here. What is the importance of the line? What kind of a line is it? An asymptote. Yes, let's put it there. In fact, it doesn't necessarily have to be a negative number, but there will be a vertical asymptote, which actually means if we start off the current right, very good, and if we, for example, if we want.Go ahead. Hey, when we cross the asymptote. In fact, when here here, I made a big deal out of it. In case your p is outside the middle range, meaning if the p is greater than the carrying capacity, or if the p is less than zero, what's going to come out of this absolute value is the same thing. Now under this condition, when you.After we take off the absolute value, now it just equal to the p over the p minus five hundred. We added in a negative sign, that's going to equal to some kind of amplitude. Then the e of the one thousand t, right? But we do notice here, even if you don't remember the detailed solution, we do know the absolute value would actually behave in exactly the same way whether the p is greater than the carrying capacity or less than zero, meaning they're combined into one.Single function, one single graph, albeit interrupted by the vertical tangent line. But when we cross the vertical tangent line, we should knowing this is a single root here would really go the other way around, and then eventually would be approaching zero. If you do the graph within the third and the first region, the graph would never get to the second region, the middle region, and the two asymptotes are still going to be the t-axis and the current capacity. So that's actually the shape of the graph, under the condition that your initial number is either greater, or like greater than the carrying or less than zero. Coming back to you. Yes or no. Ah, yes. Okay, and Elaine, are you comfortable with that again? Yeah.Alright, now the question would be, if I give you, I did, I gave you the initial condition. Now, I know it. In fact, in among many of these, okay, it depends on there are infinite many curves, and the S shapes also depends on what is your unique initial condition. But then, whatever is the point zero seven hundred, that narrows down to one curve. But then, do you include the bottom part that corresponds to the same parameter?I didn't mean highlighted red. Do we pick both branches? Again, there's no decree from heaven. Although I haven't taught you this, but you could entirely reason this out from the meaning of a picking one specific answer to a differential equation.Can we use the initial condition? We do. But using the initial condition narrows down to two pieces of the curve. Like I said, it's the same equation, just experiencing this vertical asymptote in between.Here, there are infinite many as well, but there's only one match with the upper part. They share the same coefficients.``````嗯，I'm not sure where to start. Yes, maybe hypothetically otherwise, and you're facing two options here: whether this. Branch here would call for a specific branch to match the coefficient, or it doesn't. Maybe it can be matched with some other branch. Well, the only point that we test out a solution is to plug it back into the differential equation. Correct. To test out the solution is correct or not, the two criteria. You had to plug it into the differential equation, which is this, and you know, oh, any.One of the curves we have put on the board, the infinite money, they all satisfy that difference equation, right? And the other criteria would be: Does it satisfy the initial condition? Does it satisfy that population at zero point would be seven hundred? Yeah, this white shaded curve in the bottom region does satisfy that because it doesn't even allow the x to be zero. So.This is x equals zero, and they're part of the one curve. Surely you're getting the y equal to seven hundred. So looks like everything is perfect, and it is. Well, the only question would be, what if we pair that one with actually some other branch down here, which I'm gonna actually draw that in purple. What if we pair these two? What's gonna happen in terms of a fitting into the criteria?Of the differential equation and the initial condition. I only see one purple line. See it again. I only see one purple line. Oh, there's only one purple one. I'm matching the purple with the red on the top. Well, on the top, there's no doubt that's the only branch we can pick because that's the only branch going through the initial condition, isn't it?But bottom here, even a many would actually fit into that differential equation. Because it doesn't touch the zero. Because it does touch what. Right, it doesn't touch y-axis. Correct. So that's right, but there is one curve.We can consider them as one curve. These ways defined, although they don't really have that coefficient, we have chosen.The top and the bottom are just doing the same function, right? And no, they don't look like the same. They don't have the same coffee. Guys, I do think you need to recover some intuition here. Why don't we solve it? Give me the final answer. What is your P? I also want to show you a trick.Next time you don't even go through the process again. There's no need to actually a separation of the variable, do the partial fraction, take the antiderivative. Once we know the shape of the curve here, you can directly write it out. But go ahead and solve it under this condition. So when they come out, you have to remember to add in a negative sign. At this moment, it's not hard. We just know that a is not known yet. It wouldn't be the next. Ah, for a second, is I. To go to, go into my car. Sure. Well, thank you for making it. Something car does not.Okay. Um.So, would you also do the um p over five hundred minus p equals the a e to one thousand t.Yes.嗯，然后。嗯，因为 cross multiply。 That's right.嗯。Elaine and Eddie, if you end up with the same solution, give me a confirmation. Okay.I got the same thing.Good. We also did something else to make it look simpler. Because right now there are actually two bundles of that exponential. We can actually divide that over from the top and the bottom. Divide both by the a e one thousand t. So this is going to become a five hundred. Generally, it's a carrying capacity, right? And the new denominator would actually become a one minus some kind of one over the a now.But no matter what, it's just a constant. So I'll call that a little a and e of the negative one thousand t. Please do your pencil work. Be sure to get the same conclusion. I know, in fact, you can get a general formula. This is simply going to be the PC, and this would be actually the PC that I gave you earlier. This is the other side that there's that alpha, and that's going to be the PC times alpha. It's fully generalizable.嗯。中文嗯。Yeah, I think I got the same thing. Very good. And we're gonna finish it first, and then I'll do a lot of comments. And just to answer my question now, when you graph that one, we need to use that.700 to determine what is a, and then they are one solution, and including this red highlighted branch under our there will below the t-axis, and if you're talking the purple one, and that coefficient a here would be different. But would it matter when you plug into the differential equation? If you mismatch the red one with the purple. One. Is our solution, albeit piecewise defined, but it's a legit function. It's a legit function, and it does perfectly fit into the differential equation and perfectly fits into the initial condition.``````Can I know what's going on in your head? Yes.What would be like plug in zero seven hundred into a P? Indeed. Usually the way I would eventually just intuit the whole answer would be, you know, your P. I said that we're gonna.Actually, focus on making intuitive this answer now. After we answer the previous question, but since you're talking about it, why don't we? Ah, we do know it's going to the asymptote. Surely would be five hundred if you start from a bigger than the current capacity. If you start from negative, it's going to be asymptotic going to the left. It will be approaching zero. But then whatever the asymptote here, we put it on the top. On the bottom, because we do know we're approaching it from a number higher than it. So on the bottom.It would be one minus a component here, which eventually would be approaching zero, right? And we do know it's exponential term. So we just copy that e of the ninety one thousand t, and then give it some kind of a. It's negative because remember when the t is approaching infinity, we're supposed to be approaching zero. So and there's a known constant in the front, and the way I think about it would be you think ah you think of a when the t equals zero, whatever it is complicated.In exponential form, simply vanishes, not becoming zero but becoming one. Then fundamentally, you're just asking for what do we divide into five hundred to get seven hundred? You divide by, of course, you divide by five seventh, right? So, but it's this five seventh is guaranteed to be greater than one. Sorry, I mean less than one because the reciprocal of the number greater than one. But then that means we need to subtract by two seventh from that. So the coefficient is two seventh. If you think about it, it's really eyeballable. I got that too. Ah, great, beautiful. Yeah. Alright, well then, let's finish it. Here you've got a pair of the red curves fitting into that one, but you.There are infinitely many solutions here. For example, I can actually use the initial condition into one thousand. That just means whatever this will be one half, right? You can write another one. That's the purple solution at the bottom. Come on, although each is a pair, okay? The purple one would be matching with another one on the top here, like that. And so basically, the purple one would equal to the five hundred, and then divide by one minus one half of this e of the negative ten thousand t, and it's in.The initial the y intercept would be zero and one thousand. Do we agree? Now back to the graph. We're actually talking about the two possibilities. Each one is a pair. Each one to to be called a pair simply say are saying they're honoring the same coefficient here. They have a one single expression. There's no piecewise definedness. And now I'm going to repeat to answer the question.Whether we need to keep both branches or only one? And what if I combine this red one? Oh, sorry. We combine this red one with the purple one here. So I'm highlighting that one with the purple one here. This is the purple one. Then, do they also fit into the two criteria? And justifying themselves are the answer.``````The purple one on the bottom and the red one on the top.I think we first need to see if they fit into the original equation. They do. They both do fit into the original equations. Are you right? Even in many pairs, they fit into that initial condition. I mean, the differential equation. Do they also? So fit into the initial condition then, as a pair. P zero equals seven hundred. Yep. Uh, I don't think the top purple one does. Why not? Because the purple one, the top purple one, doesn't. We're not choosing that. We're using the.Bottom purple one to pair up with the top red one, meaning you have a piecewise defined function which has two different mathematical algebraic expressions for it. This is the equation for it when t is greater than zero. Actually, no, it's greater than whatever that asymptote, which makes the denominator zero. Why don't I call that b here? And whatever this location to be, and the the second equation works to the.It is less than zero. And that's logic. You could really have a piecewise defined function. It fits into that differential equation, and it fits into the initial condition, right? Yeah. So is this a solution? It is. It is, but simply because there are infinitely many. It is, but I can change the purple into the green, and that's also.So is right because there's no constraint on the lower branch. Yeah, you could make it any anything on here as long as you kept the red one, right? Yeah, that's exactly right. So as a convention, we don't include any of these. The resultant rule thumb—it's actually more than the rule thumb—it's an axiom. And meaning you can't really argue against it. That this is sort of a convention is that please write it down when you use the initial condition to pick one.One specific solution to a differential equation. What do we call a specific solution? That means, well, before we apply the initial condition, the infinitely many curves we have solved are called the general solution. But after we have narrowed down to a specific one by satisfying an initial condition, now it's called specific solution. And the rule is, a specific solution can never include non-continuous branches. I.e., a specific solution.It must be continuous, for the reason that if it is continuous, every single component or point is determined by that initial condition. Because you move continuously from that one single point to every single other point on the curve, you move continuously. So this one single point determines the other point. But when you actually interrupt it over asymptote or discontinuity, and then we lose track, we don't know where to actually connect it to.We move to the bottom region. You can pick an arbitrary branch. For that reason, we pick none. And the way you remember it will be the solution specific solution to a differential equation is always continuous. So we only end up with with one branch. That means the solution doesn't even honor the full domain of t. The t is confined to only the t greater than b.Okay. So well, that's much ado about this bit here, but it's very important. If we're all clear on that concept now, let's come back. I'm going to actually offer you more comments so that in the future you'll be able to write that solution without having to go through this whole procedure of solving it. What do I mean? Well, first of all, we do know all of the three regions. They share the same two asymptotes, carrying capacity under zero. If we start from the middle, we could approach both continuously, one at one infinity of t. Meaning, there is going to be a one. The t is approaching positive infinity, then surely the population growth of the pound would always be approaching the carrying capacity. So for this middle region here, I would say that.Shape it easily intuitive. Well, then now we're going to translate it into algebra. Well, if eventually when the t is approaching negative infinity, the huker of it's approaching zero, what's the easiest way to do that? Well, you just put this infinity. So fundamentally, you're solving population of t now. It wants to satisfy when the t is approaching to negative infinity, the denominator would be approaching positive.The infinity, right? If so, then your p would be approaching zero. Do you guys agree with me? Can you repeat the last sentence you just said? I said, "Knowing the shape of the graph, now we know when t is approaching infinity, this pt would be approaching zero." Ah, yes, approaching zero, which actually means we know it's a fraction. The denominator would be approaching.Positive infinity那天。Yeah, which actually means it is this kind of exponential function, e to the negative something. And we know if you really want to remember what that number is, yeah, what you have to actually remember it from how we constructed it. It just said the carrying capacity, and multiplied by whatever the coefficient in the front. So that's part of the denominator to guarantee this asymptotic behavior would work, right.Yeah. On the other hand, though, so at this moment here, all the space I am leaving empty. For example, what's the real number on the numerator? That doesn't matter; it's a finite number. So this is enough to guarantee that this side works. On the other hand, it also needs to satisfy one that it is approaching positive infinity. The PT would be approaching five hundred, correct.Yeah, how do we make that happen? Well, at least we figured out the part now. Let's see. When the t is approaching positive infinity, what would this term become?嗯。T is approaching, and very with the E to nave one thousand T. Be approaching. Uh, like zero. Dina. Well done. It is approaching zero, so that term is not there anymore. We just need to be sure to be approaching five hundred. And the easiest way is to just do this. And now you're wondering, do we add or do we subtract? It doesn't really matter. We can't subtract for the reason there's no asymptote, meaning the denominator will never be zero. It won't be zero. But I said it doesn't matter because you don't even need to remember that part. At the moment here, you can just write after k.Permiting the k to be even negative. This is actually the correct answer. It's a generic solution for the middle range. It captures all the S shapes. They only differ in exactly what's the y-intercept, and that's dependent on the k. Let's see how we utilize it. If my initial condition here is actually equal to five hundred, that looks a little like what we would happen to be graphing on the board. Then can we find out what the kThe initial condition seven hundred, right? No, that's a different initial condition. I'm giving you a new one. That doesn't apply here because we're dealing with a curve in the middle region. The one you just said. When I said this.The initial condition is: Oh, I mispoke. Is it two hundred? Two hundred. As I wrote down, I'm sorry, I'm sorry if I mispoke. I didn't mean that at all. I'm on two hundred.I got K equals three hats. Yeah, I got the same thing. Good, I'm getting the same thing. To get from the five hundred to two hundred, we need to divide denominator is a two point five, and barring the one, the case one point five. I agree with you. That's it. That's how you can write down. You will find the answer. Just knowing what's the asymptote.How to determine a final coefficient using the initial condition, but you do need to know what's that coefficient on the top. That comes from the carrying capacity and also the two there. That's the alpha in the front. In case you forgot, okay, in case you really forgot now, you still have enough to go by. Well, let's suppose we don't know what that is. Ah, do ah, you know it doesn't matter. Every single.So previous step, depending on plugging in the t equal to zero. So first of all, do you understand? Do you agree that no matter even if I don't know what the number is, I'm going to leave that open and put a b here. I still got three halves, don't I? Yeah. Because plugging zero will always be one. Yes, indeed. That means you could solve the initial use the initial condition one term at a time to decide whatever the coefficients you don't remember anymore. Now we're going to actually find out.The B is. How do we still remember? Always remember, you are equipped with only two things. Okay, one is I'm gonna actually pick it in green now. One is this differential equation. If you don't have anything else, you just go back, plug all the numbers you've gotten into that one. The other one is the initial condition. So how do we find the B? Well, it's not eyeballable. You have to actually work it, but at least you could conceptual. They realize what we could do to find the bee.````````````中文嗯。中文嗯。And I know what what you guys are thinking. Always thinking of plugging it in back. I'm plugging it back into original equation.Differential equation. Yeah. Excellent, very good job. We're going to take its derivative, and see because we have the value of the p at the position zero, right? We're plugging in t equal to zero. We want to double check whether the derivative p actually matches with the differential equation. Did we not say before it always does? Yeah, but that's actually putting the b the correct b there, and now actually if.We try out unknown b, and that differential equation would guarantee it does. Alright, go ahead and do that. I think that's really good because I actually want you to take the derivative of this. But we should know it's not that complicated, right? Do we actually need to take the derivative right here?嗯。嗯。Could we remove the denominator? Yes. Into the numerator. Ah, meaning give it the negative power. Yeah. Oh, brilliant, Ivy! You really do can negative things, and that wasn't a small feat because you're giving the answer that.Improv the technique that we covered a long time ago.嗯。嗯，对，我，是，等，你，做，，do，do，you，want，me，to，write，it，down，so，this，is，the, P of the T, now, but, equal, to, five hundred, and, one, plus, three halves, of, the, E, of, the, ninety, B, T, ninety, first, and, then, I, bet, you're, doing, it, to, get, prepared, for, using, the, chain, rule.Outer function, which is simply five hundred times something negative first, and power rule applies there. And on top of it, there's an inner function that includes that exponential function, but we all know how to take that derivative of that. Yeah, that's good.So we, five hundred times one plus three has negative. The negative first power derivative. It negative how many power? Negative first. Right, but what is that derivative? Derivative of the outer function gives you what.嗯，他就是，like，that，powerful。 yeah, of course. So. Nice.One thousand times one plus three over two e na v t. Second. Where the second, the whole thing? Then the second meaning you want to include it in the parentheses.Oh, you took the derivative inner function before you even finish the derivative first function. That's actually one of the things I alerted you guys. Don't make that favorite mistake. No, you wait until you finish the very first step. Ivy, that is something native first. You simply copy that something and put the native second there. That's the derivative of the outer function, and then you multiply.Multiply the derivative of the inner function. You do not take the derivative of the inner function as you're taking the derivative of the outer function. Do you see? I, I said the, um, take the outer function's derivative and then multiply with the inner function's derivative. Right, but why did you read the negative two after you took the derivative of the inner function?Oh, I don't think I took the derivative of the inner function. Oh good. Then somehow I'm my bad. Somehow I meet I miss the hearing you. So can you give me again what just what exactly is a derivative of the outer function. 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Yes. So. Nine. One, thousand.Where does that ninety-one thousand come from? You're using the power rule. That was my answer. It wasn't yours. You're taking the power, using the power rule for this component, right? Yeah. What does the power rule say? Oh, sure, the word we over time actually. And my bad. And oh, we're dismissed. Thank you.For staying with me, I think I probably owe you guys somewhere five minutes. By the, we'll keep having good work, would you guys? Next session, we're gonna continue. By looking at how to finish up. Actually, I was talking going for a private session with Ivy. Idee Anilang, if you care to stay, you can wait until we finish the problem. Idee, you do not mind, do you? Because we're gonna continue with the same material. No. Okay, or let's.Continue now.The derivative of the inner function, right? Yeah. Which gives us. Um. Where are we gonna compete? Oh, if you want to go, that's fine. You can leave it as homework, but we're gonna compete it.Oh. Okay. Very good. Ylen, you got it. But I still want Eddie. You get it. Name:Yeah, your content, huh? Wait, wait, wait, ninety BT. No, so what is it? What is the last step? We're taking the derivative of the innermost function, which is simply this ninety BT now, right?So, may you be? Yes, did we not do it just a moment ago, Ivy? Oh no, we didn't. Sorry, I thought I did it some other way. No, no, we didn't. So, in fact, when you do the exponential function or something, and that derivative which is e to the same something, however, that something is no longer purely the independent variable. There's additional coevolution, so that something itself is a inner function. We're taking the derivative of that, which gives us an.It would be good. Well, that looks like a nightmare, but it's really not because we're going to use it to plug in the initial condition. So how do we finish it at this moment? Now we we got a derivative. And what's next? We um.Set it equal to the original position. Yes, exactly. Which should be? That's here. The two p times. What is p? Yeah, indeed, it equals two p times of five hundred minus p now.Then to determine what the B is.

Logistic 模型解与图像行为

Logistic 模型解与图像行为

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