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Point) Assuming that P > 0,a population is modeled by the differential equation dP 1.5P 5ooo dt1. For what values of P is the population increasing? Answer (in ...

Question

Point) Assuming that P > 0,a population is modeled by the differential equation dP 1.5P 5ooo dt1. For what values of P is the population increasing? Answer (in interval notation):2. For what values of P is the population decreasing? Answer (in interval notation):An EQUILIBRIUM SOLUTION is a solution that is a horizontal line with constant slope 0.3. What are the equilibrium solutions? Answer (separate by commas): P

point) Assuming that P > 0,a population is modeled by the differential equation dP 1.5P 5ooo dt 1. For what values of P is the population increasing? Answer (in interval notation): 2. For what values of P is the population decreasing? Answer (in interval notation): An EQUILIBRIUM SOLUTION is a solution that is a horizontal line with constant slope 0. 3. What are the equilibrium solutions? Answer (separate by commas): P



Answers

A population is modeled by the differential equation

$ \frac {dP}{dt} = 1.2 P \left(1 - \frac{P}{4200} \right) $

(a)
For what values of $ P $ is the population increasing?
(b) For what values of $ P $ is the population decreasing?
(c)
What are the equilibrium solutions?

In this video, we're going to be looking at population dynamics as modeled by a differential equation. So our population P is given by the differential equation at the top of your screen right here, DP DT equals 1.2 p times the quantity one minus P over 4200. So the first thing we want to figure out is where is um for what values of P is this population increasing? So um in order for our population to be increasing, we know that DP DT has to be greater than zero because we want an increasing function has a positive slope. So we want this to be a positive number. And so in order for that to be the case, That means that the quantity 1 -2 over 4200 Has to be greater than zero. Because this first part here of our differential equation is always positive, always positive. So the change in sign only comes from the second term. So if we go ahead and solve this little inequality we get one has to be greater than P Over 4200. multiply both sides by 4200. And you get He has to be less than 4200 for the population to be increased. Let's now move to decreasing. So similarly, um DP DT must be less than zero if the population is increasing because a function is it? Sorry, decreasing as a function is decreasing when its slope is negative. So um that gives us one minus P over 4200 has to be less than zero. So if we solve this inequality get one speed less than P over 4200 and we get p Greater than 1200 for population decreasing. Now we want to find equilibrium solutions. So an equilibrium solution occurs when the slope of the function is zero. So we want dP DT to equal zero. So if we go ahead and do that, let's plug in the whole equation this time. So we get from point to p times one minus P over 4200 equals zero. So we know this is already factored, so we know that um if we study to these factors zero, that will give us the values with P for which this is true. So if we do 1.2 P equals zero, Get P equals zero. And and that is our first solution there. Now, if we set the other term equal to 01 minus P for 4200 zero, we can move this over to the other side. We end up with one equals P for 4200 multiply both sides by 4200 then we get P equals 4200 for our other equilibrium solution. Um So to recap, um the function is increasing when P Is less than 4200. We have an equilibrium solution on p equals 4200, and our population is decreasing when P Is greater than 4200.

In order to find the equilibrium solution and increasing decreasing intervals. The equilibrium solutions Easiest, because we could set this equal to zero or this equal to zero. If we set the one might 1.2 p equals zero. That would occur when P equals zero. So that's one solution. Or alternatively, if we set one minus p over 4200 equals zero equilibrium is when the slope with the zero or when the derivative D P. P t equals zero. Well, then we could just move over that negative term and then set these equal multiply the 4200 over. So he is equal to 4200 is another equilibrium solution. Which makes sense, because if we substituted that value back in here, it would be one minus 4200 divided by 4200 which is one minus one, which is zero. Okay to find the other two were going to set up an interval with zero and 4200, and then we're gonna test values inside and outside of these points here. So, for instance, let's just test the value Negative one one and 4201 for instance. We're gonna plug them back into the whole differential equation here and see if they give us positives or negatives. If we plug in negative, we're going to get a negative here and then one might not a positive. So this would give us negative derivative if we plugged in. One. Get a positive times of positive so the derivative would be positive and between zero and 4200 and then 4201 and be a positive times of negative. So that would be back to a negative. Well, that immediately helps us with number or parts A and B because these air the exact intervals where we're increasing decreasing, we see that we are increasing when P is anywhere between zero and 4200. Is that a result in positive values of the derivative up top here? If we are outside of those intervals, then we'll get negative values. So those would be, for instance, if P is less than zero, or if he is greater than 4000 200 Both of those would result in a negative derivative or a decreasing P function in the end. So that's the idea with this one

In this video, we're going to be looking at population dynamics through differential equations. So the equation describing our population um that we have is right here. So we first want to figure out the carrying capacity and K value. So we know that standard form that will give us these values. Um it's going to be something like DP DT equals something times P or sorry, times one minus P over something. So we need to figure out those to some things together because this is Kay this is him and Emma's are carrying capacity. So um in order to get this one here, what we need to factor out is the value of that term. Right, so the end of with his DP DT equals point oh five p times one minus 0.1 p. And so we can rewrite point out one as one over 100. And so that gives us the final equation, DP DT equals 0.5 p times one minus p over 100. So now what we can see from that is that this 100 is m or are carrying capacity and this 0.5 is K. Now let's move on to the next part of the question. So now we have a direction field and we just want to get a little bit of intuition as to what's going on here. So where on this graph, on the slopes closest to zero, you can pretty easily see that the slope is zero at 100. Um and it approaches zero for values around it. So we know that where this looks the largest, this seems to be the largest. Uh We also can tell that they're pretty close to zero. Um had been around zero. So say you're 100. Actually, I'll just write that, don't you? 100 comma zero now where the slopes, the largest this looks are the largest one there. Um far away from these values. So exactly in the middle right here, or at 50 is where our social lives here and then up at 1 50. Their large again as well. So as far from our zero slopes as possible is where the largest and where the solutions increasing our solutions are increasing um between zero and 100. Right. And they're decreasing anywhere above that. So actually this shouldn't be a closed bracket. Will just make it an arrow. So increasing for P. Sorry, decreasing for P is greater than 100 and we're increasing for P is less than 100. All right. Next we want to sketch some solution curves. So we're giving a bunch of initial conditions um and we're going to go ahead and sketch those. So our first four initial conditions are below 100. So we have 20 forties, right about there, actually, 2020 is probably a bit higher, too. 20 40 um 16 and 80. Let's go ahead and sketch those curves. So we follow just the general trend of the slopes here. We'll see that we're going to end up with something that looks like that for our initial condition of 20. Moving on to our initial condition of 40 we'll end up with something like that. Um, 60 will be slightly different, but also follow the same general trend. And 80 will look a lot like 60. Yes. Next, let's do our ones above 100. So 1 21 40. And these are rough sketches, 20 right there. When where do you get there? This time? We are decreasing. So it's going to tend down or? Yeah. All right. So now let's talk about these solution curves just a little bit. So, um what are the commonalities and differences between these solution curves? Well if we look at them we can see that they all tend towards 100. What about some differences? Um Some are increasing, some are decreasing, some change can cavity and some don't. Right, okay, so now let's talk a little bit more about that changing can cavity. Um So which of these solutions have an inflection points and where that's what we want to now figure out. So we will um an inflection point is where a solution changes from concave up to concave down or vice versa. So if we look right away we can see that these two right here do not have inflection points and these two don't as well because our green curves, those top two green curves are always concave up and our top two black curves are always concave down. However, if we look a little closer, these ones, we can see that we're kind of gave up. Mhm. This section concave down for that section. Come came up for this section, concave down for this section. Um And so you can tell that both of the inflection points of the change in that can cavity occurs right around P equals 50. Right? So let's write that down. So um solutions with initial conditions uh P because 20 40 have election points. Uh huh P equals 15. So now finally, let's talk about equilibrium solutions pump the equilibrium solutions occur when the slope for specific y value is zero for all X values. I'm going to write those in red here and we can see that at zero where slips are always zero. So if we were to have an initial condition of zero, just get a straight line like this. And at 100 are slope is zero and would always be zero. So where we have to have an initial condition at 100 who would always be here as well. So our equilibrium solutions, our current P equals zero and P equals 100 and that is our final answer.

If you have a demand equation 7 60 minus 13 p and a supply equation for 30 plus two pink and you want to find the equal liver and point, that is where these two things will be equal. What? Okay, so I would add the 13 p to both sides, and that would give you 15 p on this side. And if we subtract 4 30 from both sides, that would give you 330. Dividing 330 by 15 gives you 22 so the equilibrium price is $22 and now we can get myself a little bit more room there. Dude, we can take the $22 and plug him into either supply or demand equation. It should be the same for both equations. So let's try it in the supply for 30 plus two times 22 is 4 74 and should be the same in the demand equation. 7 60 minus 13 times 22 and it is the same. So that would be the amount. So we have the price


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