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Fishing with quotas. In view of the potentially disastrous effects of overfishing causing a population to become extinct, some governments impose quotas that vary d...

Question

Fishing with quotas. In view of the potentially disastrous effects of overfishing causing a population to become extinct, some governments impose quotas that vary depending on estimates of the population at the current time. One harvesting model that takes this into account is$$frac{d X}{d t}=r Xleft(1-frac{X}{K}ight)-h X$$(a) Show that the only non-zero equilibrium population is$$X_{e}=Kleft(1-frac{h}{r}ight)$$(b) At what critical harvesting rate can extinction occur?Although extinction can occ

Fishing with quotas. In view of the potentially disastrous effects of overfishing causing a population to become extinct, some governments impose quotas that vary depending on estimates of the population at the current time. One harvesting model that takes this into account is $$ frac{d X}{d t}=r Xleft(1-frac{X}{K} ight)-h X $$ (a) Show that the only non-zero equilibrium population is $$ X_{e}=Kleft(1-frac{h}{r} ight) $$ (b) At what critical harvesting rate can extinction occur? Although extinction can occur with this model, as the harvesting parameter $h$ increases towards the critical value the equilibrium population tends to zero. This contrasts with the constant harvesting model in Section $3.3$ and Exercise $3.7$, where a sudden population crash (from a large population to extinction) can occur as the harvesting rate increases beyond a critical value.



Answers

Consider the differential equation
$ \frac {dP}{dt} = 0.08P (1 - \frac {P}{1000}) - c $
as a model for a fish population, where $ t $ is measured in weeks and $ c $ is a constant.
(a) Use a CAS to draw direction fields for various values of $ c. $
(b)
From your direction fields in part (a), determine the values of $ c $ for which there is at least one equilibrium solution. For what values of $ c $ does the fish population always die out?
(c) Use the differential equation to prove what you discovered graphically in part (b).
(d) What would you recommend for a limit to the weekly catch of this fish population?

We're gonna need, uh, some graphing software that's able to do, um, slope field grafts. That's not on. Not all of them can, but one, but the one I'm using can. So I'm using Geo Deborah, if you need one. Um, and it's their c A s calculator. And if you'll notice on the left here, I've already plugged in a command that gives me the slow field for our system with C at 20. And I want you to look around at the plot and see if you can tell. Is there an equilibrium solution here? So scroll in and out a little bit, and I would say it looks like there is one appearing. It's right around 500. You see, there's limited resolution on these things, so sometimes you have to move around to spot it. But right around 500 you're seeing a pretty flat line. Okay, so it looks like minus 20 Gifts Assn. Equilibrium solution. How about minus 10? Okay, so now it should be minus 10. So you're seeing there a couple interesting things here. First of all, you can probably eyeball it. It looks like there's an equilibrium solution. If I can just settle it on it. Sorry. I'm trying to get the well, Okay, maybe actually don't have to. So there's an equilibrium solution a little below 1000 and, um, right around 500. No below 500. Maybe about 200. I'm estimating a bit there. Okay. And the way you can tell, even though there are no flatlines, is if you look a little bit of 1000. There's this part where, um a bunch of all those slopes are pointing down and then immediately below them. They're pointing up. OK, so we don't from negative to positive. That means we crossed zero. So a little below 1000. There must be an equilibrium solution somewhere. It's just our resolution is not good enough to spot it. 500 is the same deal. Okay, this one's a little quite a bit below 500 actually, a little above zero. Maybe around 200. You see slopes going up and then right below them going down, they have to be flat in the middle. Okay. So even though it's not as easy to see the straight lines on this one, this direction field does show to equilibrium solutions. So we've got a little less than 20. How about a little above? Okay, a little above 20. Now, this one's very different, right? We're basically pointing down everywhere. In fact, it's so vertical. It's hard to even tell that it's down. But it is down, okay? And we can be pretty confident. I mean, we never even get close to flat that there are no equilibrium solutions here. Okay? So for 10th and 20 there were there were equilibrium solutions for 30. There weren't. Um I should I should admit, with the benefit of hindsight, I know that the cut off should be a 20. But this is the kind of experimentation you want to do in order to determine that fact. Crazy. No. Let's get to the math. That tells us we're right about this. So I'm going to write down our differential equation. This is equal, Teoh. I'm switching to fractions because I think those air almost always easier to deal with who I forgot. Api. I'm gonna use some exponents here. Scientific notation. This is 10 to the three. Instead of writing 1000 that's gonna be useful in a little bit. Minus c were interested in equilibrium, solutions and equilibrium solutions occur when the derivative is equal to zero. In this problem, we're gonna set our expression this middle one equal to zero. Because that is what it tells us. The magnitude of the derivative. And we noticed. At least it's it's beneficial to notice that, um Who I made a mistake here. Sorry about that. This is one and then this is he Okay, right? That looks better now. Sorry, I got some things jumbled in my head. No will return to the problem. So we're setting this equal to zero, and it's really beneficial to notice that this expression you have for the derivative is, in fact, a quadratic. Okay. And if you just distribute and do a little rearranging, you actually see that what we end up with is a minus A over 10 to the five. Um, he squared, plus a over 100 p minus C equals 20 Okay. And the beneficial thing is, we can plug this into our formula, right? So the quadratic formula, if you imagine this being, like, if you're used to a X squared plus b x plus C equals +20 that's the formula you're used to well, minus 8/10 to the five. That's our A, um, a over 100 that's RB. And then see, a little confusingly is minus C. Okay. And once Republic plugged that into the formula, we get negative. B lesser minus root. B squared is 64/10 to the four. That's 10,000 minus four A. C to the tube gives in the multiplication cancel out, and we actually end up with 32 over 10 to the five seen go and that's all over two A. All right? And now ask yourself basically what's important in this formula, Okay, because we don't necessarily need a lot of exact values for P. And this problem, that's not what we're asked. What we're asked are things like, When do we have no solutions versus solutions? Um, when do we have actual die off? Okay, so a lot of that is gonna have to do is just this part under the square root, right? If this part under the square root is greater than zero, well, then we have plus or minus so a positive number. We take the square root. That's another positive number plus or minus that we get to solutions. Okay, one for the plus one for the minus. If it's equal to zero. Well, plus or minus zero, that's the same thing. That's just one solution. Okay? And if it's less than zero, well, we can't. Can't take a square root of a negative number. We're dealing just with real numbers. We can't do that. So that's actually going to give us no solutions, which is entirely possible. Okay. No. When does this happen? Well, the cut off point that were interested in between, um two solution between having one solution and were more one or more solutions and no solutions is where this expression is greater than or equal to zero. Because that encompasses both the two solutions and the one solution case. Now, it turns out after you do a little bit of algebra or not even a well, you do a little manipulation that that gives you so you you add over and you end up getting see equal to 20. So if you add, if you add, is the cut off point. So you add this over Now you multiply both sides by 10 to the fifth over 32 you basically end up getting see being less than or equal to 20 or 20 being greater than or equal to see. This is the case where you have one or more solutions. No, the original OD is no longer here, but what this means kind of when you're interpreting the problem is that you shouldn't be catching more than 20 fish a day. All right, so you're somewhere between, um what the what the math tells us is between minus infinity and 20. But in reality, you know, if you're thinking about the problem, you're not gonna be catching negative fits. That would mean putting them back into the ocean. So really, you're between zero and 20 catches a day, and you're good if you have more. If you catch more than 20 fish a day, that's when you're just going to decay off. Okay? And what's happening is, Well, that's the case we plot. It's the 30 right? There are no equilibrium. Solutions just goes down. Um, and if you want a little more evidence than the graphical approach, basically it could be described this way minus C. So eventually what's gonna happen is when the sea gets too small. Well, this is a parable on the left, right. It's minus a over 10 to the five p squared, plus a over 100 p minus. See, this whole thing's a proble. And because this coefficient is less than zero, it's gonna be a Kong cave down parabola. Got a little messy there. It's gonna be so it's gonna look something like this because that's first coefficient is less than zero. So if you shift, see down for enough, you might ask, OK, why does it go to zero? Why does it decrease instead of increasing? Well, it's because the parabola is pointed down and be no zeros. There must be, is kind of a gap here, and it's gonna be negative the whole time, okay? And that means the derivative is going to be negative the whole time, okay. And that corresponds to going down. So if this were a conclave up proble, the interpretation would be that you're actually always increases. There still wouldn't be any equilibrium when they're when the quadratic gives, you know solutions, but you be increasing. But in this case, we're concave down, so we're decreasing the whole time on. If you catch more than 20 fish a day, you will kill off the entire population

Here. PN represents a fifth fish population after and years. And that population is modelled by this formula over here. Okay, so for part A, we will show the following. If the sequence converges, then the limit it's and goes to infinity is either zero or be minus a. So let's go ahead and verify this fact solution. So let's just go ahead and supposed So we're supposing that it converges delicious. Call that limit l now using the given formula for piano up here. Let's take a limit on both sides of this equation as n goes to infinity. So we're applying the limit on both sides. So this becomes and this can be simplified now, too. The quadratic. Okay, so now at this that l equals zero is a solution to this equation over here. However, if l is not equal to zero, then we can divide it to obtain. And there are two solutions to this quadratic zero for B minus. And that was what we wanted to establish for parting. So if it converges the limit, it either dies off at zero or it stabilizes that B minus a, which is a constant. Okay with that said, Let's go on to the next page for a heartbeat. So this is where we like to show the following inequality. So it's quite and give a solution For this we have P n plus one, by definition or by the Riker JH in formula Given for p n. We can write this now. Let's divide top and bottom by a A number is just one, and then we get P n over a. Now I can just go ahead and ignore that denominator. The reason I can do this, we're dealing with positive numbers. Here's when the beginning A was a positive fish Population PM has to be not negative. So this is bigger than or equal to one or in this case, just equal to one. So that justifies this inequality here and that resolves the party because this is what we wanted PM plus one and then be over a PM And here this was You justify this. It's okay. Let me go on. I'll need some more room here, so let me go on to parse even on the next page. So here we like to use party to show that if a is bigger than be then Lim api and Ghost zero as n goes to infinity. Okay, so one way to show this is the following. So if the Siri's PN converges, then by the test for diversions we have the limited pn zero. So let's just go ahead and try to establish this fact here and then we'LL finish the problem. So let's try the ratio test for this using part see in mind because in part of seeing remember, she sees me from part B. We had key to the n plus one. It was less than be over, eh, Tien? So now we try the ratio test here. Since we're just dealing with positive numbers, we could drop the absolute value. This is B over a incense now and part see, we're assuming is larger than be that will imply be over in less than one so that the limit of Ki n plus one over p n is less than one. So that means that the Siri's converges by the ratio test and as we mentioned the beginning by the test for divergence, that will imply that the limit of P and zero and that's also what we what we wanted to establish a policy. Now there's one more party here, so we'LL go on to the next page for party. Now we're assuming is less than a B and there's a few things to show here. So first we want to show that if peanut is less than a B minus a, then we'd like to show that the sequence is increasing. And we'd like to also show that Pien is between zero on B minus eight Then on the other hand, if peanuts bigger than be minus a then we'd like to show these things The PM is decreasing and dance p and is larger than B minus a And then finally we'LL make a conclusion at the end that it is less than B and the limit of PN is b minus. Okay, so what's that? Improving this so and black thes two parts appear this worldview to parts and then two parts for the green as well. So this will take a few moments. So let's go ahead and start proving this. So we were getting so we're assuming this. But we were also given that peanut was bigger than zero, so we actually have zero less than Pena and then Liston d minus a. So now we would like to go ahead and let's show this one first. Let's do that by induction. We already have the bass case here. Bass case corresponding to an equal zero. Now, let's go ahead and uses for the induction. So we'd like to go ahead and suppose that it's true for PN and then show that would be induction. Okay, so now let's recall the following. We have tea and plus one by definition. Now, I'm just gonna do a trick up here in the numerator. And then I could rewrite this as the following. So now going on to the next page. So this was our assumption here for the inductive step at eight to both sides. Go ahead and divide. Bye, baby, he reckons, multiply both sides by the reciprocal zor football sides. And here let me put a negative motivated by what we had in the previous page. So this gives us again going back to the previous page here and then using our newest information. And that's exactly what we wanted. We wanted to show that if it was true for n that It was also true for en plus one and let's see here and by definition we know that zero will also be less than just by definition, of p m in the fact that a, B, t and R all positive. So this shows one of the parts we still have another parts ago. So now we have the following we now we want to show that it's increasing. So let's just recall that definition again. And then Now let's you not the family when we have what we already showed. So using the same type of trick is before and there we go. We see that PM is increasing sequence. So this takes care of the first part that we mentioned that was in black. And now similarly, there's two parts to show for the green steps. So this was if now we assume peanuts bigger than be minus a and so this is what we'd like to show. And then we'LL do it by induction again. Already we have the bass case up here. So now let's go ahead and recall the definition of P n plus one using our trick that we've used several times. So now since PM is bigger than be minus a. This would be our inductive step inductive hypothesis. It's at eight of both sides. Divide by eighty football sides and then multiply both sides by a negative on the next page. And so this will give us by previous work and then buy our latest inequalities. And once again, there we go. Now we have it for N plus one. Now we want to show that it's decreasing. So this was also supposed to be in green, but I moved on to a new page, so we'd like to show this. So we have add that aids of the other side using some algebra here. So this will give us that p n plus one bp on over a plus pn the shrink we wants than PM And so that's what it means to be decreasing. And so we had one more party here, and this one is to show that the following So if so, I'm getting a little sloppy. You're sorry about that. There was three parts. The party. That's why I'm using these down to you. So if peanuts less than a B minus in, then by what we showed this is positive, increasing and bounded above bye. Part by the first part of Heartbeat. We showed these things, and so his conversion bye theorems won't monotone sequence there and then also since the key and is bigger than zero and it's increasing. It can converse zero because it's increasing. So by partying we showed that the limit was either zero or B minus, and if it can be zero, it has to be B minus eight. Similarly, if peanut is bounded below, then buy this party. What we also showed is that pee and still positive. But this time it's decreasing. This is what we did in green and bounded below. Might be money, so it's converging also, by the twelve months a sequence there and then sense, um, Tien is bigger than be minus a. We have to have that the limit of P and bigger than equal city minus ing and which is bigger than zero so that the limit of PN is not equals zero and again by party. The limited it's non zero has to be a B minus, and that resolves the problem

Mhm. Okay, So for the first part of this problem, we want to find all equilibrium. The first thing to do is to conceptually think about what this means. So if the dynamics of this system is described by differential Equation, which is the case here, then equilibrium could be estimated by setting a derivative or all derivatives equal to zero. So in this case, there's just one derivative term this d n over GT term. So in order to find all equilibrium, what we want to do is set this equal to zero. Then, from there, we look at the right half of this equation, and what we can plug in for these values depends on what's given in the question. So the question tells us that our is equal to two and that K is equal to 1000. So those are the values that we can plug in on this right side. Then, from here, we're able to solve. So now it's just algebraic Lee going through and solving this expression. There's different steps you could take to do this. I would first distribute this too and get to minus n over 500 times and minus age times and and then I see here that both of the terms in this expression have a factor of em. So I'm going to pull this factor out to the front and rewrite this expression as and times two minus and over 500 minus h. And so now I've rewritten this expression as a product of two factors with one factor being n and the other being to minus and over 500 minus age from here. In order to find equilibrium, values or expressions, we set both of these factors equal to zero. So the first one is pretty easy and is equal to zero, and that is one of our equilibrium for this expression. Then we set this second factor equal to zero to minus on over 500 minus each and such since and is the term for the population number. This is what we're going to isolate and get alone. So I'm going to add and over 500 to both sides and then multiply both sides of this expression by 500 to get an alone. And so that gives me N is equal to 1000 minus 500 h. So this is your second equilibrium solution for part A, then moving on to part B two b specifies that we want to look at this non zero equilibrium that we just found. So this is the N equals 1000 minus 500 h on. We want to find the critical value of H, which is the point at which the population will go extinct. So we want to think about this conceptual e if this is the equilibrium solution we're looking at and this is the actual portion that represents what the population is going to be to find the point at which it's going to go extinct, we want to find the point at which this expression for the population is going to drop below zero. Right, So it's going to be less than or equal to zero or the point at which zero is going to be greater than or equal to this population. From here we sell for H, so I would add 500 each to both sides. Divide both sides by 500 you end up getting that age is greater than or equal to two. So your answer for this would be a value off for each that's equal to two would be that critical value. When it goes greater than this, the population goes extinct thin. The last component for this problem is part C. We now wanna look at where this non zero equilibrium expression is going to be locally stable. So here it's kind of the flipside of what was discussed in B. Now we wanna look at where we're going to have a positive population value. If you were looking at it graphically, it would actually be the H values orphan on an X and Y axis you can think of as the X values for which the graph would be above the X axis. So you could think of these values being representative of that. But if we wanna algebraic Lee solved, it's going to be again thinking about it. Conceptually, we want to think about this portion that represents the population when it's going to be greater than zero, or when zero is going to be less than this expression. So again, I'm just going to sell for each by adding 500 to both sides. Divide both sides by 500 you get that it's locally stable when this Each value is less than two, and so that would be your answer for part C.

Okay, So for this question, we know that D and E t is equal to you. Our times one minus and over k times and minus h. So we know that our is equal to two so pleasant values in we end up with ours equal to 21 minus Kay's equal to 1000 and H is equal to 100. So we said this equal to zero to solve for equilibrium points So 100 is equal to two and minus two and squared, divided by 1000. Factoring out and we end up with two minus, we contract for it to end. So we end up with to end times one minus and over 1000 is equal to 100 so n is equal to 50. And you also know that one minus and over 1000 misty equal so and over 1000 has equal negative 99 or an is equal to negative 99 times. 1000 right to determine whether they're stable or unstable. We define a function G that is equal to two times That's he that is equal to you two times one minus and over 1000 and minus 100 soul for G prime and and Check whether GE Primal 50 and G Private connected 99,000 whether they are greater than or less than zero.


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