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Problem 20 Hard Difficulty

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?

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Calculus 2 / BC

Calculus: Early Transcendentals

Chapter 9

Differential Equations

Section 4

Models for Population Growth

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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

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James Stewart

Calculus: Early Transcendentals

Catherine Ross

Caleb Elmore

Michael Jacobsen

Joseph Lentino

This textbook answer is only visible when subscribed! Please subscribe to view the answer

Calculus: Early Transcendentals

Catherine Ross

Caleb Elmore

Michael Jacobsen

Joseph Lentino

James StewartCalculus: Early Transcendentals

Catherine Ross

Caleb Elmore

Michael Jacobsen

Joseph Lentino

James Stewart

Calculus: Early Transcendentals