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# Suppose that a population develops according to the logistic equation$\frac {dP}{dt} = 0.05P - 0.0005P^2$where $t$ is measured in weeks.(a) What is the carrying capacity? What is the value of $k?$(b) A direction field for this equation is shown. Where are the slopes close to 0? Where are they largest? Which solutions are increasing? Which solutions are decreasing?(c) Use the direction field to sketch solutions for initial populations of 20, 40, 60, 80, 120, and 140. What do these solutions have in common? How do they differ? Which solutions have inflection points? At what population levels do they occur?(d) What are the equilibrium solutions? How are the other solutions related to these solutions?

## a) $k=0.05, M=100$b) The slopes are closest to zero near P=0 and P=100. The largest positive slopes are near P=50, and the largest negative slopes are near or above P=150. Solutions are increasing for positive values of P that are less than 100. Solutions are decreasing for P greater than 100.c) Solutions for $P_{o}=20,40$ Have inflection points at $P=50$d) $P=0$ or $P=100$

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Differential Equations

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Thuth G.

July 14, 2021

Suppose the population of the world was about 6.2 billion in 2000. Birth rates around that time ranged from 35 to 40 million per year and death rates ranged from 15 to 20 million per year. Let's assume that the carrying capacity for world population is 20

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Hey, it's glorious that when you married here. So we have part A. We're gonna find the carrying capacity, which we see as equal to DP over. DT is equal to 0.5 p minus Sierra 0.0 five p square. Now we get the logistic model equation for finding the carrying capacity, which is equal to 0.5 p. Times one minus 0.1 p. We got a DP over DP, which is equal to 0.5 See Time's one minus p over 100 and we get Kate to be equal to 0.5 or M is equal to 100 now. For part B, we see that the slopes are closest to zero near P as equal to zero and P is equal to 100. The largest positive soaps is on P is equal to 50 and the largest negative is when P is near or above 1 50 We see that it's increasing for positive values, or pee that are less than 100 are decreasing when the P value is greater than 100. Next, we're going to do some graphing So we see that piece of zeros, the initial population. So all the solutions they're gonna be going to P is equal to 100. So when ti, as time goes infinity, we see the equilibrium solution of the differential equation. It's gonna be 100. So you see, that piece of zero is bigger than 100. These are decreasing and piece of zero is smaller than 100 are increasing. You know that the slopes will be the biggest one p of equal to 50 and between zero and 100 um, the negative of the slope will increase as p goes to above 100. And when t is equal to zero and he is equal to 100. Since these are the equilibrium solutions, a slope will be close to zero. And then for the inflection points, we see that it is when Pete it is a P is equal to 50 for one piece of zero is equal to 20 or 40. So this is the graph of 20 looks. This is a rough sketch. So we do this, it's gonna be 20 and it just goes like this. And then this is for 40 you see that it's getting more of like a humpback shape and then we're doing 60 and 80 so compared to 40 60 it's more has two more of the curve here. The beginning and then 60 and 80 pretty much look similar, but when we hit 1 20 we go all the way from up and it kind of goes down now for part D. We know we know that DP over d t has to be equal to zero. So we get DPS over DT. We get this part from part, eh? 0.5 p minus 0.0 five. P square is equal to through and we get P is equal to thorough and P is equal to 100. Since we get 0.5 over a 0.5 when P is equal to zero, it means that that there were no populations in the beginning, so at any time there's none. And when P is equal to 100 it shows that the it will reach equilibrium, which means that the number of deaths are equal to the number of birds. When the population becomes 100

#### Topics

Differential Equations

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