WEBVTT mathematics/ap-calculus-ab/hovasapian
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Hello, welcome back to www.educator.com, and welcome back to AP Calculus.
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Today, we are going to be talking about a particular differential equation.
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There are several that we could have chosen from but we decided to go with population growth.
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We are going to be discussing the standard and the logistic equations for population growth.
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Two differential equations, the standard we are just going to go through to mention it.
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It is the logistic equation the one that we really want to concentrate on.
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Let us jump right on in.
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There are many differential equations that exist that try to model how populations grow and decay.
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Let us see, I think I will go ahead and work in blue.
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There are many differential equations that try to describe population growth and decay.
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I will just put growth/decay.
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The simplest of these is something called the standard model or the natural model standard equation, the natural equation.
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Whenever you hear the word model, essentially, you are talking about a different equation
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or series of differential equations that try to describe the situation.
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Model is just a fancy word for that.
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The simplest of these equations, the simplest of these is the standard model or natural model.
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It says the following, it says that the rate of change of a population
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is directly proportional to the size of the population at the moment.
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The size is of the population at the moment.
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Symbols is this, dp dt, the rate of change of the population.
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The rate at which the population changes, dp, per unit change in time dt.
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The derivative is a rate of change or rate of change is just a derivative.
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The rate of change of a population is directly proportional to,
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that means some constant × the size of the population at the moment p.
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That is it, this is a model.
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The differential equation is this.
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We seek a function p(t), notice we did not use dy dx, dp dt, change in population versus the time.
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Here, the independent variable is the time, the dependent variable is the population.
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We seek a function p(t), that will predict what the population will be at any future time.
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We want to solve this differential equation.
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Once we have the solution, this p(t), we want to compare it to data
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that we have collected on how populations grow and decay.
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If the equation matches the empirical data, the real life data, your model is a good model.
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If it is not, then we have to go back and modify the model or come up with a different model altogether.
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That is essentially all that we do in science.
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As we go into the lab and we collect a certain amount of data, we try to describe it mathematically with some model.
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In other words, a set of differential equations, 1 equation, 2 equations, 3 equations, whatever it is that we need.
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We try to see if it matches, once we solve the differential equation, we try to see if it matches the actual data.
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That is it, that is really all science comes down to.
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Sometimes the data itself will, oftentimes, the data itself will tell us, gives us clues on how the model should look.
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It is sort of back and fourth between the data, the model,
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until we have refined the model so much that it actually describes what we observe in the real world.
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Once that happens consistently, we call that a theory.
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That is all that is going on in science.
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Let us go ahead and see what we can do.
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Let us to solve this particular equation.
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We have that the rate of change of population with respect to time is proportional to the size of the population of the moment.
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It looks like this is a separable equation.
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Let us go ahead and separate variables here.
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We are going to end up with dp/p = k × dt.
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Sure enough, it is separable, that t is on one side and the p is on the other.
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Since it is separable, all we have to do is integrate both sides.
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The integral of dp/p is the natlog of p.
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I do not need the absolute value sign here because a population is always going to be positive.
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P is always positive, it is just the natlog of p is equal to kt + c.
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This is our general solution of this differential equation.
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Let us go ahead and change the form and make it look like an exponential.
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Let us exponentiate both sides.
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We end up with p is equal to e ⁺kt + c.
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Kt + c, the exponent, this is just e ⁺kt × e ⁺c.
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Same base at the exponents.
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E ⁺c is just some number is just some number based on c.
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I will call the whole thing, this whole thing is just some constant.
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We will call it a.
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What we have is the population at time t is equal to some a × e ⁺kt.
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The standard model predicts an exponential growth of a population.
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Another reason why it is actually called the natural model because it is based on the natural logarithm.
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This is the natural model.
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Basic, first, approximation model, we predict exponential growth.
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Let us go ahead and say a little bit more about this.
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This equation is usually accompanied by some initial values, initial conditions.
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In other words, we know what a population is at any given moment, when we start to collect the data.
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In the year 2000, the population of New Haven, Connecticut was 400,000, whatever it was.
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We have some initial data.
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That is our initial value problem.
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This equation is usually accompanied by a set of initial conditions, usually, just one initial condition.
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In other words, the population at time 0, whatever our 0 time happens to be, is going to equal some number.
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I will call it p sub i, it is the initial population of our group.
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This is just the initial population at time 0.
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The initial value problem is nothing more than the differential equation dp dt = kp.
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We have p at time 0 = p sub i.
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We have the general solution.
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We just solved it, we have the general solution.
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That is just p(t) is equal to ae ⁺kt.
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Let us see if we can use this initial value to find what a is going to be.
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p(0) which means put 0 in for x, or in this case t, is equal to a × e ⁺k × 0.
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They are telling me that p(0) which is equal to this, when I put it in the equation is equal to p sub i.
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e ⁺k0 is e⁰ it is equal to 1.
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a × 1 = the initial population, a is equal to the initial population.
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Our solution is, the population at time t is equal to the initial population that I start with at time 0 × e ⁺kt.
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K is the constant, in this case.
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For different values of k, there is going to be different degrees of growth.
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Let us do an example, this is our general equation.
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At 1:00 pm on June 15, a biologist counts a bacterial population of 100 in a culture.
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At 11:00 pm on the same day, the culture has grown to 187 bacteria.
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Assuming a standard growth model, how much bacteria will there be at 1:00 pm on June 18, 3 days later?
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Three days later, 1 pm, 1 pm, let us see what we have got.
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P(t), we are assuming a standard model so our solution is p(t) = π e ⁺kt.
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We know what the initial population is, it is 100 bacteria in the culture, at = 0 which is 1:00 pm on June 15.
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We have to choose a unit of time for t.
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Are we going to work in days, minutes, seconds, years, months.
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June 15 and June 18, that is three days.
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1 pm, 1 pm, that is 11:00 pm.
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Let us work in hours, actually, it looks like.
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We choose to express time in hours.
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I think it will be the best.
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We know what the initial population is, we need to find what k is.
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Once we find what k is, then we can go ahead and plug in a value of t to find what p(t) is.
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The π we know, in this case, we have one constant π, we have another constant k.
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We need to find both of these.
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The π, we know that is just 100.
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We have p(t) is equal to 100 × e ⁺kt.
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Let us go ahead and solve for k.
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Let us see what I have got.
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P(t) is equal to, I already got that, I do not have to repeat that.
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At 11:00 pm, the population is 187 bacteria which means,
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What that means is that, p(10), 1:00 pm is t = 0.
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11:00 pm is 10 hours later, at t = 10.
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P(10) = 187, I can put these values into this equation to see what I get for k.
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P(10) which is equal to 100 e ⁺k × 10 is equal to 187.
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I solve here, e ⁺10k is equal to 1.87, 10k is equal to the natlog of 1.87.
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I get k is equal to the natlog of 1.87 divided by 10.
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The natlog of 1.87 divided by 10 which is equal to 0.06259.
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Our equation becomes, population of t is equal to 100, the initial population, × e raised the power of 0.06259 t.
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1:00 pm on June 18, three days later exactly, three days is 72 hours, three days later exactly, t = 72 hours.
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Therefore, what we are looking for is p(72).
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P(72) is equal to 100 × e⁰.06259 × 72.
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When I do that, I get 9,060 bacteria.
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Let us look at this equation.
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Let me go back to blue here.
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I have population = an initial population × e ⁺kt.
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There are four parameters here.
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There is final population, the initial population, the growth constant, k and t.
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There are four parameters in this equation.
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If you have any three of them, you can find the fourth by just rearranging, solving, that is the whole idea.
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If you have any three of them, you can solve for the fourth.
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The nature of the problems that you are given are going to be such that they allow you to find three of them.
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You do not know which three, that is why no two problems actually look the same.
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You are going to have to reason out which three you are going to find.
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Once you found those three, you will find the fourth.
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That is it, that is all that is going on here.
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Let us go back to blue.
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A much improved model for population growth.
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The standard model, it is not bad, it works to a certain degree.
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But just from your experience, you know that an exponential growth, something cannot just keep growing exponentially.
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A population is not just going to keep growing and growing and growing.
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People leave, people die, people are born, people move in.
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Not to mention the fact that there is only a certain number of amount of resources that are available.
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A particular environment can only support a maximum number of people.
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You know just intuitively that a population will tend to grow but that eventually it will start to level off,
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once you actually reached what we call the carrying capacity of that particular environment.
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Whether it is a petri dish or whether it is a city or a country or whatever it is.
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A little bit better, a lot better, a much improved version and much improved model
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for population growth and decay is the logistic model or the logistic equation.
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This particular differential equation looks like this.
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It says the rate of change of a population is not just directly proportional to the population.
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It is directly proportional to the population × 1 – what the population is/ c, where c is the carrying capacity of the environment.
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In other words, carrying capacity of an environment is the maximum number of people that the environment can support.
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Like a room, when you go to a room and it says maximum occupancy 250, only 250 people can be in that room.
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If there is more than 250, people have to leave.
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If there is less than 250, people will keep coming in and the population will increase until you reach 250.
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That is it, it is just exactly what it sounds like, the carrying capacity.
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Where c is the carrying capacity which is the number of individuals,
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whether that would be people, bacteria, whatever, the number of individuals an environment can carry and support.
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Now this is our differential equation, the logistic model.
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Let us take a look at what is going to happen here.
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Those of you who would actually go on in your scientific careers, you are going to run through the differential equations.
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Oftentimes, you actually do not need to solve them.
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All you need to do is analyze them and look for what we call qualitative behavior.
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You are going to look at a different equation, you may not be able to solve the equation
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but you can extract a lot of information from just looking at the equation
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and seeing if you can get this qualitative information from it.
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Usually that will be enough for you to answer whatever question you are trying to answer.
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In this particular case, let us see what happens,
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if the population at any given moment is less than the carrying capacity.
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If p is less than c, this p/c is less than 1.
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1 - a number less than 1 that means this 1 - p/c term is positive.
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This implies that dp dt, if this is positive, this is going to be positive which means the population is going to grow.
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Dp dt will be positive.
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A positive rate of change means something is growing, which means that p is actually growing.
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The population will grow.
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It just confirms our intuition, population will grow until it levels off at c.
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In which case, when the population reaches c, this becomes 1 – 1, it is 0.
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There is no more population growth, it just levels off.
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The rate of change of population is 0.
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It is not growing, it is not decaying, it just stops right there.
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This is qualitative information, if the population at any given moment is actually greater than the carrying capacity,
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then this term 1 - p/c is negative.
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This implies that dp dt is negative.
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A negative rate of change means that the dependent variable, the population in this case, is declining.
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It is negative which means the population will decrease until it levels off at the carrying capacity.
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We already have some idea of what the solutions are supposed to look like graphically.
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You are going to see a population increase and eventually it is going to level off at the caring capacity.
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If the population is greater than the carrying capacity, the population is going to decline until it levels off at the carrying capacity.
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We already have some idea what the graph should look like, what the solution should look like.
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That is why this qualitative information is very important.
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Let us go ahead and solve this equation.
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Let us solve this initial value problem.
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The initial value problem is dp dt is equal to k × p × 1 - p/c.
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P(0) is equal to p₀ or p sub i, initial population, some number.
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This is separable, this differential equation is separable.
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May not look like it, but it actually is.
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Let us go ahead and do, here is what I'm going to do.
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I’m going to write this equation and I’m going to multiply this out.
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I’m going to write this as dp dt is equal to k × pc - p²/ c.
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That is just this, multiplied out, common denominator multiplied out.
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Now what I’m going to do is I'm going to move the dt up here.
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I’m going to move c, it is just a constant.
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I’m just going to bring everything here this way.
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What I end up with is, once I separated out and hopefully you can take care of this.
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I’m not going to go through the entire process pc - p² dp is equal to k dt.
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Now the p’s are all on one side, constant does not matter, it is just a constant.
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The t’s are on one side.
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I can go ahead and I can integrate.
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Now integrating a rational function.
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This c, I’m going to decompose this by partial fraction, left side, when I integrate this.
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The difficulty here is not the separation, the difficulty here is the actual integration.
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pc - p², I'm going to factor the denominator c/p × c - p
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which is equal to some a/p, partial fraction decomposition + b/ c – p.
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Now I have got, let me go to the next page here.
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Let me write this again.
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I have got c/ p × c - p is equal to a/p + b/ c – p.
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Multiply, multiply, partial fraction decomposition, I get c is equal to a × c - p + bp.
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c is equal to ac - ap + bp.
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c is equal to ac + -a + b × p.
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Therefore, ac 1c, this coefficient is equal to the coefficient from here.
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a is equal to 1, that takes care of the a.
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I have got -a + p is equal to 0 because there is no p term here on the left,
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which implies that a is equal to b which implies that b is also equal to 1.
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Our c/ p × c - p is equal to actually 1/ p + 1/ c – p.
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This is our a and this is our b.
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Therefore, the integral of our c/ p × c - p dp is equal to the integral of 1/ p dp + the integral of 1/ c – p dp,
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which is equal to the natlog of p - the natlog of c – p.
00:30:08.800 --> 00:30:20.400
Therefore, what we have is c – p.
00:30:20.400 --> 00:30:34.100
This is the left side of the integral.
00:30:34.100 --> 00:30:43.400
Let us go ahead and rewrite what is it that we actually did.
00:30:43.400 --> 00:30:53.500
We had the integral of c/ p × c - p = the integral of k dt.
00:30:53.500 --> 00:31:10.500
We just did that integral, that is ln of p - ln of c - p = the integral of this is just kt + some constant.
00:31:10.500 --> 00:31:12.700
I will go ahead write out constant, I do not want to use the word c
00:31:12.700 --> 00:31:17.700
because I do not want you to confuse it with this c right here.
00:31:17.700 --> 00:31:35.300
Over here, we have the natlog of p/ c - p is equal to kt + a constant.
00:31:35.300 --> 00:31:47.800
I’m going to flip this and I’m going to write this as the natlog of c - p/ p = -kt - the constant.
00:31:47.800 --> 00:31:49.700
I hope that make sense.
00:31:49.700 --> 00:31:54.500
Log of a/b is log(a) – log(b).
00:31:54.500 --> 00:31:57.500
If I flip this, it just takes the negative sign out here.
00:31:57.500 --> 00:32:01.700
I move the negative sign to this side, now I have got that.
00:32:01.700 --> 00:32:06.500
And now I exponentiate both sides.
00:32:06.500 --> 00:32:22.300
I end up with c - p/ p is equal to e ⁻kt × e ⁻some constant.
00:32:22.300 --> 00:32:32.100
This constant, I'm just going to call a.
00:32:32.100 --> 00:32:48.500
I’m going to separate this out, this is going to be c/p - p/p which is 1 = a × e ⁻kt.
00:32:48.500 --> 00:33:05.100
I have got, this is going to end up giving me c/p = a × e ⁻kt + 1.
00:33:05.100 --> 00:33:09.300
When I solve this for p, I’m going to move the p up here, move this down here.
00:33:09.300 --> 00:33:21.700
I'm left with p is equal to c which is the carrying capacity not the constant, = a which is some constant e ⁻kt + 1.
00:33:21.700 --> 00:33:26.800
This is my solution to the logistic equation.
00:33:26.800 --> 00:33:31.900
My population is equal to the carrying capacity divided by this thing,
00:33:31.900 --> 00:33:39.300
some constant a × the exponential, some growth constant t + 1.
00:33:39.300 --> 00:33:42.900
Let us go ahead and deal with the initial value.
00:33:42.900 --> 00:33:52.400
Initial value, it is said that the population at time 0 is equal to some initial population.
00:33:52.400 --> 00:34:01.100
Therefore, our p(0) is equal to c.
00:34:01.100 --> 00:34:10.100
We use our equation ae ⁻kt × 0 + 1.
00:34:10.100 --> 00:34:15.400
They tell me that it is equal to p(0), the initial population.
00:34:15.400 --> 00:34:28.500
e⁰ is just 1, what I have here is c / a + 1 = p(0).
00:34:28.500 --> 00:34:43.700
I rearrange this, c/ p(0) = a + 1.
00:34:43.700 --> 00:34:51.200
I get a is equal to c/ p(0) – 1.
00:34:51.200 --> 00:35:04.600
This is one version of it or when I do a common denominator, I get a = c – p(0)/ p0.
00:35:04.600 --> 00:35:12.100
Our equation that we solved was the carrying capacity/ a × e ⁻kt + 1.
00:35:12.100 --> 00:35:20.300
This value of a, I get it by taking the carrying capacity - the initial population divided by the initial population.
00:35:20.300 --> 00:35:29.300
I have a way of finding a.
00:35:29.300 --> 00:35:33.100
Let us go back to blue.
00:35:33.100 --> 00:36:05.300
Our initial value problem dp dt = kp × 1 - the population/ c, where c is the carrying capacity.
00:36:05.300 --> 00:36:12.200
p(0) = p₀, some number, which is the initial population.
00:36:12.200 --> 00:36:14.200
It gives the following.
00:36:14.200 --> 00:36:25.900
Our solution is, the carrying capacity divided by a × e ⁻kt + 1,
00:36:25.900 --> 00:36:33.000
where a is equal to the carrying capacity - the initial population/ the initial population.
00:36:33.000 --> 00:36:38.400
This is the solution to our problem.
00:36:38.400 --> 00:36:44.000
What happens as t goes to infinity, as time just grows?
00:36:44.000 --> 00:36:59.400
Let us analyze this solution, what happens when t goes to infinity?
00:36:59.400 --> 00:37:08.000
When t goes to infinity, e ⁻kt goes to 0.
00:37:08.000 --> 00:37:17.700
a × e ⁻kt goes to 0.
00:37:17.700 --> 00:37:26.000
As this goes to 0, the population goes to c.
00:37:26.000 --> 00:37:31.300
As time increases, the population goes to c, the carrying capacity, exactly what we said.
00:37:31.300 --> 00:37:40.300
A population will grow and then it will level off at the carrying capacity.
00:37:40.300 --> 00:37:46.500
Let us actually see what the solutions look like.
00:37:46.500 --> 00:37:50.400
I have a couple of cases.
00:37:50.400 --> 00:37:58.200
Case 1, what if the initial population is less than c?
00:37:58.200 --> 00:38:05.700
Let us just pick some numbers, let us say our initial population of 100 and a carrying capacity of 2000.
00:38:05.700 --> 00:38:09.300
2000 is the maximum population in this environment.
00:38:09.300 --> 00:38:15.600
We have a formula for a, a is the carrying capacity - the initial population/ the initial population
00:38:15.600 --> 00:38:24.500
which is 2000 - 100/ 100, that is going to equal 19.
00:38:24.500 --> 00:38:32.300
Case 2, it is where the initial population is actually bigger than the carrying capacity.
00:38:32.300 --> 00:38:33.800
Let us pick some numbers.
00:38:33.800 --> 00:38:41.500
Let us do an initial population of 3000, where the carrying capacity is 2000.
00:38:41.500 --> 00:38:53.000
We have a, a is equal to the carrying capacity - the initial population
00:38:53.000 --> 00:39:03.400
divided by the initial population which is equal to 2000 - 3000/ 3000.
00:39:03.400 --> 00:39:08.200
It is going to be equal to -1/3.
00:39:08.200 --> 00:39:27.300
Our case 1, our equation, p is going to equal the carrying capacity 2000 divided by a which is 19 × e ⁻kt + 1.
00:39:27.300 --> 00:39:41.900
In our case 2, our population is going to be the carrying capacity 2000 divided by a which is -1/3 e ⁻kt + 1.
00:39:41.900 --> 00:39:50.500
What I'm going to do is I’m going to graph that equation and I'm going to graph that equation for different values of k.
00:39:50.500 --> 00:39:53.200
k is the nature of the growth.
00:39:53.200 --> 00:39:57.700
The value of k is something that we deduce from data that we have collected.
00:39:57.700 --> 00:39:59.200
Let us take a look at what these look like.
00:39:59.200 --> 00:40:04.500
I have taken a particular case, a particular initial population, a particular carrying capacity.
00:40:04.500 --> 00:40:10.200
A particular initial population and a particular carrying capacity.
00:40:10.200 --> 00:40:19.700
Here is what they look like.
00:40:19.700 --> 00:40:31.300
Initial population of 100 for different values of k, k(0.5), k(0.7), k(0.9).
00:40:31.300 --> 00:40:38.900
They all start at an initial population, they start to grow, and then they level off at the carrying capacity.
00:40:38.900 --> 00:40:41.900
The carrying capacity is 2000.
00:40:41.900 --> 00:40:43.700
For different value of k, same thing.
00:40:43.700 --> 00:40:47.300
The behavior is the same, it is the nature of the growth.
00:40:47.300 --> 00:40:52.300
This definitely is a much better model for how populations actually behave.
00:40:52.300 --> 00:40:55.900
They will eventually start to grow, that is resources start to deplete,
00:40:55.900 --> 00:41:01.900
the environment can only support so much so they achieve a maximum carrying capacity.
00:41:01.900 --> 00:41:09.500
This is for when the initial population is less than the carrying capacity.
00:41:09.500 --> 00:41:13.700
The initial population is 100, carrying capacity was 2000.
00:41:13.700 --> 00:41:19.000
There is going to be initial growth and it is going to the exponential growth, but then it is going to level off.
00:41:19.000 --> 00:41:21.100
That is what is happening here.
00:41:21.100 --> 00:41:24.700
For the case where the initial population is greater than the carrying capacity,
00:41:24.700 --> 00:41:32.200
the initial population of 3000 with a carrying capacity of 2000, it is going to just decay.
00:41:32.200 --> 00:41:34.300
For different values of k, that is all it is.
00:41:34.300 --> 00:41:39.700
These different graphs are just different values of k, that has to do with the nature of the environment.
00:41:39.700 --> 00:41:46.300
It is going to decline until it reaches the carrying capacity and it is going to level off.
00:41:46.300 --> 00:41:47.900
That is what is happening here.
00:41:47.900 --> 00:41:52.100
Let us go ahead and do an example.
00:41:52.100 --> 00:41:57.700
Solve the following initial value problem then use the solution to find the time,
00:41:57.700 --> 00:42:03.700
when the population reaches 75% of its carrying capacity.
00:42:03.700 --> 00:42:09.800
In this particular case, dp dt = 0.065.
00:42:09.800 --> 00:42:12.800
This is our k, they gave us the k, in this case.
00:42:12.800 --> 00:42:18.700
P1 p/c, this is our carrying capacity.
00:42:18.700 --> 00:42:20.500
This is our c, our carrying capacity.
00:42:20.500 --> 00:42:23.800
The initial population is 120.
00:42:23.800 --> 00:42:26.500
We have a lot of information here.
00:42:26.500 --> 00:42:31.600
We have our k which is equal to 0.065.
00:42:31.600 --> 00:42:36.600
We have an initial population equal to 120.
00:42:36.600 --> 00:42:41.400
We have a carrying capacity of 1500.
00:42:41.400 --> 00:42:43.300
We know what our solution looks like.
00:42:43.300 --> 00:42:48.100
First of all, let us find what a is.
00:42:48.100 --> 00:42:51.400
We know what the solution is already, we solve this differential equation.
00:42:51.400 --> 00:42:57.900
It is equal to c/a × e ⁻kt + 1.
00:42:57.900 --> 00:43:01.200
Let us go ahead and find what a is.
00:43:01.200 --> 00:43:10.100
I know what a is, a is equal to carrying capacity - initial population/ initial population.
00:43:10.100 --> 00:43:17.900
It is equal to 1500 which is the carrying capacity - 120 which is the initial population/ 120.
00:43:17.900 --> 00:43:21.800
I get an a value of equal to 11.5.
00:43:21.800 --> 00:43:25.400
This is the value that I put in there.
00:43:25.400 --> 00:43:30.300
I have a k value of 0.065, I put that value in there.
00:43:30.300 --> 00:43:38.300
My equation is p is equal to the carrying capacity which is 1500, and that goes here.
00:43:38.300 --> 00:43:41.000
I have already found most of the parameters here.
00:43:41.000 --> 00:43:59.800
1500 divided by 11.5 × e⁻⁰.065 t + 1, that is my solution.
00:43:59.800 --> 00:44:11.600
I want to find the time when the population reaches 75% of its carrying capacity.
00:44:11.600 --> 00:44:22.900
75% of c is just equal to 0.75 × 1500 which is equal to 1125.
00:44:22.900 --> 00:44:28.600
The equation that I’m going to solve is, when the population is 1125.
00:44:28.600 --> 00:44:30.900
I want to solve for t.
00:44:30.900 --> 00:44:43.800
It is 1500/ 11.5 × e⁻⁰.065 t + 1.
00:44:43.800 --> 00:44:51.200
I want to solve for t.
00:44:51.200 --> 00:45:07.000
I’m going to do 1125 × 11.5 e⁻⁰.065 t + 1 = 1500.
00:45:07.000 --> 00:45:21.200
I get 12937.5 e⁻⁰.065 t + 1125 = 1500
00:45:21.200 --> 00:45:36.700
e⁻⁰.065 t is equal to 0.028986 subtract 1125 divided by the 012937.5.
00:45:36.700 --> 00:45:41.800
I end up with this, take the natlog of both sides.
00:45:41.800 --> 00:45:50.700
The natlog of this, the natlog of that, I end up with a final value of t = 54.5.
00:45:50.700 --> 00:45:51.600
I do not know the units.
00:45:51.600 --> 00:45:54.900
The units could be seconds, hours, minutes, days, years, whatever.
00:45:54.900 --> 00:45:58.700
It is just 54.5 time units.
00:45:58.700 --> 00:46:01.700
Let us see what this looks like.
00:46:01.700 --> 00:46:03.200
This is what this looks like.
00:46:03.200 --> 00:46:13.700
An initial population of 120, the carrying capacity of 1500, it is right there.
00:46:13.700 --> 00:46:24.400
It starts to grow, at the t value of 54.5 which is here, I labeled it as 54.48,
00:46:24.400 --> 00:46:33.900
I hit a population of 1125 which is 75% of my carrying capacity, which happens to be up here.
00:46:33.900 --> 00:46:42.600
That is all that is going on here, I hope that made sense.
00:46:42.600 --> 00:46:47.600
Let us see here, what else can I do?
00:46:47.600 --> 00:46:51.500
One last thing to round things out.
00:46:51.500 --> 00:47:00.500
There is one final concept that comes up in your problems, may or may not come up, but just in case it does.
00:47:00.500 --> 00:47:11.800
One final concept that can come up.
00:47:11.800 --> 00:47:16.300
We speak about the rate of change of population dp dt.
00:47:16.300 --> 00:47:20.200
Sometimes we speak of the relative rate of change of population.
00:47:20.200 --> 00:47:23.100
There is a difference, here is what they are.
00:47:23.100 --> 00:47:36.900
Relative rate, you need to know how to turn that word problem into some symbolic form.
00:47:36.900 --> 00:47:48.200
Relative rate, this is the topic that can come up now for population.
00:47:48.200 --> 00:47:56.500
Growth rate is what we already know, that is just dp dt.
00:47:56.500 --> 00:48:00.200
Growth rate, it is the rate of change of the population per unit change in time.
00:48:00.200 --> 00:48:02.300
It is the derivative, dp dt.
00:48:02.300 --> 00:48:13.300
Relative growth rate or relative rate of growth or relative rate, you might see it that way too.
00:48:13.300 --> 00:48:26.800
Relative growth rate, it is equal to dp dt, the growth rate divided by the population at the time.
00:48:26.800 --> 00:48:33.400
Or you can do it this way, 1/p dp dt.
00:48:33.400 --> 00:48:36.600
The growth rate is dp dt, the normal derivative.
00:48:36.600 --> 00:48:42.900
The relative growth rate is the normal derivative divided by whatever the population happens to be.
00:48:42.900 --> 00:48:52.400
In other words, it is how fast the population is changing relative to how many there are in the population.
00:48:52.400 --> 00:48:56.000
Relative just means divide by that number.
00:48:56.000 --> 00:49:02.000
It is a ratio that you are actually coming up with.
00:49:02.000 --> 00:49:05.100
In terms of equations, it is going to look like this.
00:49:05.100 --> 00:49:21.700
For the standard model, we said that dp dt is equal kp.
00:49:21.700 --> 00:49:28.000
The rate of growth is directly proportional to the population at the time.
00:49:28.000 --> 00:49:35.500
Relative growth version is this.
00:49:35.500 --> 00:49:46.900
Relative growth version is 1/p × dp dt is equal to k.
00:49:46.900 --> 00:49:52.900
In words, this one is going to say, the relative growth rate of the population is a constant.
00:49:52.900 --> 00:49:54.100
They are the same equation.
00:49:54.100 --> 00:49:59.100
The only difference between this equation and this equation is I actually just divided by p and I brought it over here.
00:49:59.100 --> 00:50:05.400
In words, it is a same equation, there is no difference.
00:50:05.400 --> 00:50:11.700
The solution that we got for the standard model is going to be the same solution here.
00:50:11.700 --> 00:50:13.700
We are still going to separate variables and integrate.
00:50:13.700 --> 00:50:15.900
The only difference is it is going to be worded differently.
00:50:15.900 --> 00:50:23.100
Here they are going to say the rate of change in the population is directly proportional to the population.
00:50:23.100 --> 00:50:28.400
Here they are going to say the relative rate of change in the population is constant.
00:50:28.400 --> 00:50:34.400
You are going to say the relative rate of change divided by the population is equal to a constant.
00:50:34.400 --> 00:50:36.800
That is all that is happening here.
00:50:36.800 --> 00:50:46.900
Growth rate, relative growth rate, just divided by whatever the dependent variable is.
00:50:46.900 --> 00:50:54.700
It is the same equation just worded differently.
00:50:54.700 --> 00:51:03.900
You have to be aware of that, just worded differently.
00:51:03.900 --> 00:51:06.000
Thank you so much for joining us here at www.educator.com.
00:51:06.000 --> 00:51:07.000
We will see you next time, bye.