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ZK
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Problem 1 Medium Difficulty

A population grows according to the given logistic equation, where $ t $ is measured in weeks.
(a) What is the carrying capacity? What is the value of $ k? $
(b) Write the solution of the equation.
(c) What is the population after 10 weeks?
$ \frac {dP}{dt} = 0.04P (1 - \frac {P}{1200}), P(0) = 60 $

Answer

a) $k=0.04$ and carrying capacity is 1200
b) $P(t)=\frac{1200}{1+19 e^{-0.04 t}}$
c) 87

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Emre C.

September 13, 2021

Estimating the height of a building a meter standing 30 ft from the base of a building measures the elevation angle to the top of the building as 75°. How accurate is the angle so that the error in the estimated height of the building is less than 4 % sho

Video Transcript

Hello. So today the problem that we're going to do involves a growth rate differential equation which is given by DP over TT. And this is going to equal kay, which is our growth Read a relative growth read times P and Pete is gonna be our population s O. This is the change in publishing of the change in time and then this is multiplied by one Linus P. Holder and and M is going to be our carrying capacity. So the first question in our particular case, the equation that were given his point zero four times p times one minus p over twelve hundred. So in our particular case, the first question asked, What is the carrying capacity? So the carrying capacity for us if we compare, um, this M, which is the carrying capacity with twelve hundred we get that are carrying capacity on DH, is equal to twelve hundred and this is our carrying capacity. And then for the second part of pardon A, we have what is the value of K Ah, and again Okay, if we compare that with our general equation, OK, which is our relative growth raid is going to be point zero four. So this first one is really just knowing the general equation, knowing what the variable stand for on DH then on and then just equating those. So for the second part, right, the solution of the equation. So for this part, what do we want Teo on? We want to solve this differential equation. So in section nine point three, it talks about solving this type of differential equation on, and this is what's called a separate ble differential equation. And the reason that it's separable is because on the right side of the equation year, we have point zero for and then pee times one minus p over twelve hundred. And this is considered sufferer ble because we khun separate in tow. One function that's just in terms of P on multiplied by another functions just in terms of team. So you can see that there's actually no tease in the right side of the equation, which is fine because we could just use point zero four is our equation for tea. It just means that we we can separate all the teas from the peas on DH two two equations where were multiplying one equation It just has teas and one equation that just as peas. So we can do that here because we don't necessarily. We don't have any tears s o in this type of situation, What we want to dio. So the first thing I want to dio is I'm going tio, find common denominators and and and get this second term in the prentiss. He's ah, simplified A little bits is his point zero four times p. And now this is thiss one is going to be twelve hundred over twelve hundred, so we're gonna have twelve hundred minus p over twelve hundred. So now what we conduce now. So this this means, implies a CZ. We can now move all the terms, so move this D t it the bottom over the right hand side of the equation and remove all the terms with P and to the left hand all the terms with tea, which we don't have any terms with you to the right and this side of the security of the equation. So this is going to become what this is going to become, so I'm gonna multiply by both sides by twelve hundred. Um, this is gonna be twelve hundred and then divide both sides by twelve hundred minus p. So this is going to be over twelve hundred minus p on this, the multiplying we're keeping the dp on the side and this is going to be equal. We only have point zero for Oh, so we forgot a term here term we forgot squint and be on the bottom here on that term is what we need to keep this p on the bottom just then still implied. Sign them. So now this is equal that point zero four and we just have DT on the side. Remember, we moved the DT over. So now the idea is we want to integrate both sides the left hand side in terms of p, the right hand signed in terms of D. So now when we do that, um, let's see what happens here in this. I would be nice to this race a little faster, but that's okay. So what do we want to do? I want to integrate the right side in terms of DP Once in a great side, the left side, in terms of DP, want to integrate the right side in terms of tea. So now when we integrate the left side, we're gonna have to do Ah, partial fractions here. So the way that we do that is ah, off to the side. Um, so let's add a new page in. So we're going toe of twelve hundred, and this is going to be over. Um, it was a p Times twelve hundred minus p. Right. And we want to set this equals we want to break this in tow up into two factions on We want to find some letter, some some number. Her expression, eh? That's over this first term p. So a overpay plus some B um, it appears that be axe is a good question. Um, I think it's just being I believe that it's going to be over. Just be Yes, yes, his inhibitors and being. And then this is over twelve hundred nine s p. So now what do we want to do? We wantto combine these two fractions on the right hand side. We need to get common denominators. We do that by just multiplying the two denominations. So this left hand side is going to equal. Ah eh, Times twelve hundred in this pea plus b times p and this is going to be over a common denominator of P times twelve hundred minus p. So now we want to do with partial fractions. We're going to set thes numerator is equal. You can think about it if you want, by multiplying through by the denominator cancels out on both sides, eh? So what we're going to get because we're going to get twelve hundred on the left hand side and this is going to equal. So now I'm going to distribute is a in So we're gonna have twelve hundred, eh? And now this is going to be minus a times p on DH plus being times p. So now let's move all the terms, the right hand side of the equation. So we're going to get the zero is equal to minus twelve hundred plus twelve hundred. Hey, I'm gonna put this imprint disease and I'll explain why in a second and then now we have minus a P plus BP to free factor at API. From that, that's gonna equal plus peen times B minus. Hey, the idea with this is we have this equation on the right hand side equal to zero. Right? So he said each of the coefficient equals zero. You can think about the left hand side. Is this a CZ being a Merrill right here you can think of this is like zero close Ah, p Times zero. And we're kind of setting these coefficients equal. So when we do that, our first equation is Is this minus twelve hundred plus twelve hundred? Hey, and we want to set this equals zero, then saw in this equation the next line, What we're going to get turning at twelve hundred is equal to twelve hundred, eh? And then dividing by twelve hundred and more sides, we get that A is going to equal one. From the second equation, we set the coefficient of pea to the first power equal to zero. So now from that equation, we're going it be minus A is equal to zero. So from that equation, what we get yet that b equal to a which is equal to one. So in this top equation here, we're now going tohave thiss where a is equal to one and b is also equal The one So if we go back Tio r integration here. Um, this is going to be equal, Tio. What? To the unit Girl of one overpay plus one over twelve hundred minus p dt. And then the right hand side. We could just integrate. That s So what is that gonna be? That's going to be pointing zero for T on. There's one thing to know. We're going toe. So typically we're integrating. Both sides were going to get two constants of integration if we wanted to. We could move both of those. Those constants, right? Anytime we we do it indefinitely, we were gonna have a constant. Because when we take the droop of that Constance gonna be zero. So, really, what we're getting is a family of functions that that we could get. You know what? When we're evaluating this interval, but we're getting these two different constants, so we can kind of just combine them. Maybe in the one constant. Ah, you can think about it. But, you know, if I had to see one on the right and then when I do this in around the laughter, get another C too. You know, I could just subtract it over and get C two months he one called a C three. The point is, we were really on anyone constant here, and we're going to once we solve. Once we want to resolve this, um, this differential equation, we're gonna have a constant, and we're going to use our initial condition toe to solve that constant you. So let's do this. Now we're integrating one overpay. So what is that? That's going to be the natural log of That's the value of P. And then we're integrating than twelve hundred, minus P s O. That is going to be the natural log of P. We were actually goingto have a little issues. I I guess technically this should be a user, but were too pretty easy use of war. You know, it's just the fact of negative one. So we we know that this is going to be something along the lines of natural log of twelve hundred minus p. And then let's take the drivel of that. And what do we get? We get one over twelve hundred minus p and then we have to multiply by the truth inside, which is negative one so that negative one is going to cancel this negatively impact plus one off her twelve hundred minus B. So that's a good trick with integration, right after you're done integrating. Ah, you can take the derivative and you should end up back with what you had before you. You you integrated s So now it's at our constant re certain. So plus, C and this is going to equal point zero for tea. Yeah, right. Um, so right. So now I guess we we have to solve or, um for P. Do we really want to do that? Apparently, Yes. Apparently. Radio wanted to get that. Um so right, Linds. Student. So what are we gonna have? We combine this night. These two natural logs. When I subtract, that's the same his division. So I'm gonna have the natural long. Um, So I guess the other thing to note his one Ah p his population. Um right. So I was just wondering if we could get rid of these absolute values. So we we know that population camping negative per se. Ah, we don't know that pee is less than twelve hundred. Ah, at least not necessarily from what I see on, right. So? So I guess we can't necessarily conclude that. So this is gonna be a over P over twelve hundred minus p on DH said natural log adapts of value. Here, let's move the sea over the other signs. We have point zero for tea. There's his plus C. Um, so now I think we're going to need a new sheet. He, um So now what we have we have the natural log of P bye twelve hundred minus p. And this was equal. Teo, tell a check in turns here, point zero forty. See? So let's take e razed to the both science. Um right. So then we'll haveto the natural log so hee to the natural log of this side friend. And when I have to the natural log of P over twelve hundred minus p. This's going Teo Teo, the point zero for tea plus the right this e to the natural log that those two were going to cancel each other out. So what we're going to be left with is the absolute value of P over twelve hundred minus p. This is going to equal he to the point zero forty. Plus he right, So now what do we need to do I'm going to assume for now that pee is less than twelve hundred. Um, if Pierre's bigger than twelve hundred, what's gonna happen? This this, this numerator when I take that's the value's going toe foot. Ah, so let's for now assume it's too soon that he is strictly less than twelve hundred. If it wanted to be equals twelve hundred, because then we'd have zero on the Dominator them. So if we assume that we dropped ups of values on now, let's push to the other side. The twelve was Multiply the twelve hundred Miles Peter the other side. Then we'LL get P equals twelve hundred Linus P times he to the point zero for tea plus C. Now we multiply through and we get P equals twelve hundred times E to the point zero forty plus C minus minus P he teo point zero for tea plus C. Now let's move all the terms with Pito left hand side of the equation on get P plus p times E to the point zero forty Plus he Yes, I should have probably solve for this constant integration a while ago because now I'm separated out into multiple visas that's a little bit annoying. Um, whatever. Williams still do it. It's not a big deal. We might We might even go back to an older equation to do that power down. Yeah. Anyways, so now we have This is equal to twelve hundred, he to the point zero for tea less than a c. So now we want to do is run a factor out a peen on the left hand science. This is going to be p times one plus he to the power of point zero forty plus he and this equals twelve hundred times e to the point zero for tee. Let's see. So, no, I think we're going to a new page, but we want to divide by this term, multiplying by P on the left hand side. So we're going to end up with p is equal Tio on the way. And so twelve hundred. We have twelve hundred, he to the point zero. There's not enough in there. No one's you for tea. Plus he this is going to be over one close. Uh, what was it? He to the point zero for tea plus scheme. So once we saw for our constant innovation C. We can plug that in. And this will be our solution to the differential equation. Ah, let's news. Ah, this equation you come back here. Maybe here soap equals twelve hundred minus p. So p e clothes twelve hundred minus p Think time Z to the point zero for a plus. So now we want to do is we want to use our initial conditions. We know that p At times zero this is equal to what, zero to sixty. So let's plug in zero for tea and plug in sixty for p and then solve this. So then what we have If sixty solver first see, the constancy is this is sixty equals Prentiss, Here's twelve hundred minus sixty. And then this is times me so point zero four times zero zero zero Plus I guess I'll just write it zero plus see to receive power. So now we have sixty. So this is going to be eleven. Forty. So we have sixty weaken divide by that eleven forty So divided by eleven. Forty this is going to be equal tio he to this sea. So now we get what we get If we take the natural audible sides. We get that our constancy is going to equal the natural log of six. So I guess six over one fourteen or, um, you could do three over fifty seven. It's X three over fifty seven. So now let's go ahead and plug that in. Look that back into our original equation for P. So I guess let's let's on the side here. Before we do that, just say what, Right? So welcome, we dio weaken dio this s o yeah, that's zone a new page. So we have our equation. Let me this we copy it for P is equal tio twelve hundred a hundred times he the point zero for tea Plus the this is over one Plus he the point zero fourteen. Let's see. So we want to do what we know what c is. He is a natural, long something ofthree over fifty seven. Right? So let's take that and let's see what happens. And we So either the point zero for tea. Plus he what is this going to be equal? Tio. So when were multiplying to two things with the same base like every multiplied friends ends X squared plus extra fourth we add the exponents so we could do the same thing in reverse. We have either the point zero. Forty plus c that's going to equal lied to the point zero forty times e to the c again. If you go back in the other direction, you can see that that's gonna be equal, cousin, you're adding the exponents. So now this is going to be equal to e to the point zero for tea and now e to the C This is going to be e to the natural log ofthree over fifty seven. But again, the e and the natural log you're going to cancel out. We're just going to be left with three over fifty seven times e to the point zero fourteen. So and see if I had a Yeah. So plug this in our calculator. So on the in the numerator, we're going to get what we're going to get P is equal to, and it's going to be twelve hundred times three over fifty seven's. What? Siri off works nicely. Twelve hundred times three hundred fifty seven. It's not that nice. Um, so yeah, yes. Two fifty seven five three. Yes, it must. It does. So fifty seven divided by three is in every nineteen. This is actually one over nineteen. So we're gonna have twelve hundred over nineteen, because again, this is going to be the one over nineteen times E to the point zero forty. So this is times E to the point is, you're for tea. Divided by what? Here? Funded by one plus one over nineteen times E to the point zero for team. That is going to be our solution to the differential equation. So this is the ends her to be the last problem. The last part of the problem is going to be to find this his parts he a job for for this is going to find the population after ten weeks. So we want p of ten. That's it. P of turn your equation for P was what there is this here. So we have twelve hundred over nineteen. There were multiplying that by ye to the point zero for times t just turn on dh. Then we're goingto be dividing. This the denominator we have won plus one over nineteen times. He to the point zero for times ten right, So now plugging us and the calculator. What are we going to get? So see twelve hundred Time's weaken Tio to the point. Fourth power. Um, then divided by we bring that nineteen to the bottom, and it's going to be in nineteen. Plus you to the point forth on DH. I get that. This is around eighty seven point three. Sick one to fire a round S O P. S zero zero with the sixty. Ah, So does this make sense? Um, it's the least, doesn't it? Yes. So there's no that's those Other than that, it's nothing that suggests to me that this wouldn't make sons friend. So I hope you guys find thought this was useful. I think it was a little bit longer. Maybe then, ah ah, originally intended. I do think that that sometimes when people teach this in different classes, they they ah, for instance, kind of memorized solutions the differential equations, Um, you know, general solutions so that we could have we could have potentially saved time by doing that, as opposed to solving the for the coefficients and the partial factions. I think it's useful ways the first couple times that you do these types of problems that that you go through all the steps. So I hope you thought thought that I hope that you thought that it I hope that you found this useful on DH. I will gladly see you later.