AP Calculus AB Population Growth: The Standard & Logistic Equations

Hello, welcome back to www.educator.com, and welcome back to AP Calculus.0000

Today, we are going to be talking about a particular differential equation.0005

There are several that we could have chosen from but we decided to go with population growth.0010

We are going to be discussing the standard and the logistic equations for population growth.0014

Two differential equations, the standard we are just going to go through to mention it.0019

It is the logistic equation the one that we really want to concentrate on.0024

Let us jump right on in.0028

There are many differential equations that exist that try to model how populations grow and decay.0032

Let us see, I think I will go ahead and work in blue.0041

There are many differential equations that try to describe population growth and decay.0043

I will just put growth/decay.0069

The simplest of these is something called the standard model or the natural model standard equation, the natural equation.0074

Whenever you hear the word model, essentially, you are talking about a different equation0082

or series of differential equations that try to describe the situation.0086

Model is just a fancy word for that.0089

The simplest of these equations, the simplest of these is the standard model or natural model.0092

It says the following, it says that the rate of change of a population0116

is directly proportional to the size of the population at the moment.0135

The size is of the population at the moment.0151

Symbols is this, dp dt, the rate of change of the population.0164

The rate at which the population changes, dp, per unit change in time dt.0178

The derivative is a rate of change or rate of change is just a derivative.0183

The rate of change of a population is directly proportional to,0188

that means some constant × the size of the population at the moment p.0191

That is it, this is a model.0195

The differential equation is this.0200

We seek a function p(t), notice we did not use dy dx, dp dt, change in population versus the time.0203

Here, the independent variable is the time, the dependent variable is the population.0219

We seek a function p(t), that will predict what the population will be at any future time.0222

We want to solve this differential equation.0250

Once we have the solution, this p(t), we want to compare it to data0254

that we have collected on how populations grow and decay.0259

If the equation matches the empirical data, the real life data, your model is a good model.0264

If it is not, then we have to go back and modify the model or come up with a different model altogether.0271

That is essentially all that we do in science.0276

As we go into the lab and we collect a certain amount of data, we try to describe it mathematically with some model.0279

In other words, a set of differential equations, 1 equation, 2 equations, 3 equations, whatever it is that we need.0286

We try to see if it matches, once we solve the differential equation, we try to see if it matches the actual data.0292

That is it, that is really all science comes down to.0298

Sometimes the data itself will, oftentimes, the data itself will tell us, gives us clues on how the model should look.0301

It is sort of back and fourth between the data, the model,0309

until we have refined the model so much that it actually describes what we observe in the real world.0312

Once that happens consistently, we call that a theory.0322

That is all that is going on in science.0325

Let us go ahead and see what we can do.0328

Let us to solve this particular equation.0330

We have that the rate of change of population with respect to time is proportional to the size of the population of the moment.0333

It looks like this is a separable equation.0342

Let us go ahead and separate variables here.0344

We are going to end up with dp/p = k × dt.0347

Sure enough, it is separable, that t is on one side and the p is on the other.0351

Since it is separable, all we have to do is integrate both sides.0354

The integral of dp/p is the natlog of p.0358

I do not need the absolute value sign here because a population is always going to be positive.0363

P is always positive, it is just the natlog of p is equal to kt + c.0367

This is our general solution of this differential equation.0374

Let us go ahead and change the form and make it look like an exponential.0378

Let us exponentiate both sides.0382

We end up with p is equal to e ⁺kt + c.0384

Kt + c, the exponent, this is just e ⁺kt × e ⁺c.0391

Same base at the exponents.0398

E ⁺c is just some number is just some number based on c.0400

I will call the whole thing, this whole thing is just some constant.0415

We will call it a.0421

What we have is the population at time t is equal to some a × e ⁺kt.0426

The standard model predicts an exponential growth of a population.0433

Another reason why it is actually called the natural model because it is based on the natural logarithm.0441

This is the natural model.0445

Basic, first, approximation model, we predict exponential growth.0447

Let us go ahead and say a little bit more about this.0454

This equation is usually accompanied by some initial values, initial conditions.0457

In other words, we know what a population is at any given moment, when we start to collect the data.0462

In the year 2000, the population of New Haven, Connecticut was 400,000, whatever it was.0469

We have some initial data.0476

That is our initial value problem.0477

This equation is usually accompanied by a set of initial conditions, usually, just one initial condition.0481

In other words, the population at time 0, whatever our 0 time happens to be, is going to equal some number.0510

I will call it p sub i, it is the initial population of our group.0518

This is just the initial population at time 0.0524

The initial value problem is nothing more than the differential equation dp dt = kp.0540

We have p at time 0 = p sub i.0549

We have the general solution.0555

We just solved it, we have the general solution.0560

That is just p(t) is equal to ae ⁺kt.0566

Let us see if we can use this initial value to find what a is going to be.0573

p(0) which means put 0 in for x, or in this case t, is equal to a × e ⁺k × 0.0579

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.0594

e ⁺k0 is e⁰ it is equal to 1.0602

a × 1 = the initial population, a is equal to the initial population.0610

Our solution is, the population at time t is equal to the initial population that I start with at time 0 × e ⁺kt.0614

K is the constant, in this case.0627

For different values of k, there is going to be different degrees of growth.0631

Let us do an example, this is our general equation.0639

At 1:00 pm on June 15, a biologist counts a bacterial population of 100 in a culture.0647

At 11:00 pm on the same day, the culture has grown to 187 bacteria.0653

Assuming a standard growth model, how much bacteria will there be at 1:00 pm on June 18, 3 days later?0659

Three days later, 1 pm, 1 pm, let us see what we have got.0669

P(t), we are assuming a standard model so our solution is p(t) = π e ⁺kt.0677

We know what the initial population is, it is 100 bacteria in the culture, at = 0 which is 1:00 pm on June 15.0689

We have to choose a unit of time for t.0702

Are we going to work in days, minutes, seconds, years, months.0704

June 15 and June 18, that is three days.0707

1 pm, 1 pm, that is 11:00 pm.0711

Let us work in hours, actually, it looks like.0714

We choose to express time in hours.0720

I think it will be the best.0736

We know what the initial population is, we need to find what k is.0745

Once we find what k is, then we can go ahead and plug in a value of t to find what p(t) is.0755

The π we know, in this case, we have one constant π, we have another constant k.0764

We need to find both of these.0768

The π, we know that is just 100.0769

We have p(t) is equal to 100 × e ⁺kt.0774

Let us go ahead and solve for k.0794

Let us see what I have got.0798

P(t) is equal to, I already got that, I do not have to repeat that.0800

At 11:00 pm, the population is 187 bacteria which means,0811

What that means is that, p(10), 1:00 pm is t = 0.0834

11:00 pm is 10 hours later, at t = 10.0842

P(10) = 187, I can put these values into this equation to see what I get for k.0847

P(10) which is equal to 100 e ⁺k × 10 is equal to 187.0862

I solve here, e ⁺10k is equal to 1.87, 10k is equal to the natlog of 1.87.0877

I get k is equal to the natlog of 1.87 divided by 10.0895

The natlog of 1.87 divided by 10 which is equal to 0.06259.0905

Our equation becomes, population of t is equal to 100, the initial population, × e raised the power of 0.06259 t.0922

1:00 pm on June 18, three days later exactly, three days is 72 hours, three days later exactly, t = 72 hours.0948

Therefore, what we are looking for is p(72).0976

P(72) is equal to 100 × e⁰.06259 × 72.0979

When I do that, I get 9,060 bacteria.0991

Let us look at this equation.1001

Let me go back to blue here.1007

I have population = an initial population × e ⁺kt.1020

There are four parameters here.1029

There is final population, the initial population, the growth constant, k and t.1032

There are four parameters in this equation.1050

If you have any three of them, you can find the fourth by just rearranging, solving, that is the whole idea.1061

If you have any three of them, you can solve for the fourth.1068

The nature of the problems that you are given are going to be such that they allow you to find three of them.1086

You do not know which three, that is why no two problems actually look the same.1095

You are going to have to reason out which three you are going to find.1101

Once you found those three, you will find the fourth.1104

That is it, that is all that is going on here.1106

Let us go back to blue.1113

A much improved model for population growth.1116

The standard model, it is not bad, it works to a certain degree.1120

But just from your experience, you know that an exponential growth, something cannot just keep growing exponentially.1125

A population is not just going to keep growing and growing and growing.1131

People leave, people die, people are born, people move in.1136

Not to mention the fact that there is only a certain number of amount of resources that are available.1144

A particular environment can only support a maximum number of people.1150

You know just intuitively that a population will tend to grow but that eventually it will start to level off,1156

once you actually reached what we call the carrying capacity of that particular environment.1162

Whether it is a petri dish or whether it is a city or a country or whatever it is.1167

A little bit better, a lot better, a much improved version and much improved model1172

for population growth and decay is the logistic model or the logistic equation.1185

This particular differential equation looks like this.1204

It says the rate of change of a population is not just directly proportional to the population.1207

It is directly proportional to the population × 1 – what the population is/ c, where c is the carrying capacity of the environment.1213

In other words, carrying capacity of an environment is the maximum number of people that the environment can support.1230

Like a room, when you go to a room and it says maximum occupancy 250, only 250 people can be in that room.1236

If there is more than 250, people have to leave.1246

If there is less than 250, people will keep coming in and the population will increase until you reach 250.1248

That is it, it is just exactly what it sounds like, the carrying capacity.1254

Where c is the carrying capacity which is the number of individuals,1257

whether that would be people, bacteria, whatever, the number of individuals an environment can carry and support.1265

Now this is our differential equation, the logistic model.1284

Let us take a look at what is going to happen here.1289

Those of you who would actually go on in your scientific careers, you are going to run through the differential equations.1294

Oftentimes, you actually do not need to solve them.1300

All you need to do is analyze them and look for what we call qualitative behavior.1302

You are going to look at a different equation, you may not be able to solve the equation1309

but you can extract a lot of information from just looking at the equation1312

and seeing if you can get this qualitative information from it.1317

Usually that will be enough for you to answer whatever question you are trying to answer.1322

In this particular case, let us see what happens,1325

if the population at any given moment is less than the carrying capacity.1328

If p is less than c, this p/c is less than 1.1333

1 - a number less than 1 that means this 1 - p/c term is positive.1345

This implies that dp dt, if this is positive, this is going to be positive which means the population is going to grow.1358

Dp dt will be positive.1366

A positive rate of change means something is growing, which means that p is actually growing.1376

The population will grow.1382

It just confirms our intuition, population will grow until it levels off at c.1385

In which case, when the population reaches c, this becomes 1 – 1, it is 0.1401

There is no more population growth, it just levels off.1410

The rate of change of population is 0.1414

It is not growing, it is not decaying, it just stops right there.1415

This is qualitative information, if the population at any given moment is actually greater than the carrying capacity,1420

then this term 1 - p/c is negative.1431

This implies that dp dt is negative.1438

A negative rate of change means that the dependent variable, the population in this case, is declining.1446

It is negative which means the population will decrease until it levels off at the carrying capacity.1458

We already have some idea of what the solutions are supposed to look like graphically.1483

You are going to see a population increase and eventually it is going to level off at the caring capacity.1488

If the population is greater than the carrying capacity, the population is going to decline until it levels off at the carrying capacity.1494

We already have some idea what the graph should look like, what the solution should look like.1501

That is why this qualitative information is very important.1505

Let us go ahead and solve this equation.1513

Let us solve this initial value problem.1521

The initial value problem is dp dt is equal to k × p × 1 - p/c.1532

P(0) is equal to p sub 0 or p sub i, initial population, some number.1542

This is separable, this differential equation is separable.1550

May not look like it, but it actually is.1559

Let us go ahead and do, here is what I'm going to do.1564

I’m going to write this equation and I’m going to multiply this out.1568

I’m going to write this as dp dt is equal to k × pc - p²/ c.1570

That is just this, multiplied out, common denominator multiplied out.1585

Now what I’m going to do is I'm going to move the dt up here.1590

I’m going to move c, it is just a constant.1594

I’m just going to bring everything here this way.1595

What I end up with is, once I separated out and hopefully you can take care of this.1598

I’m not going to go through the entire process pc - p² dp is equal to k dt.1605

Now the p’s are all on one side, constant does not matter, it is just a constant.1616

The t’s are on one side.1619

I can go ahead and I can integrate.1622

Now integrating a rational function.1625

This c, I’m going to decompose this by partial fraction, left side, when I integrate this.1630

The difficulty here is not the separation, the difficulty here is the actual integration.1639

pc - p², I'm going to factor the denominator c/p × c - p1645

which is equal to some a/p, partial fraction decomposition + b/ c – p.1655

Now I have got, let me go to the next page here.1667

Let me write this again.1679

I have got c/ p × c - p is equal to a/p + b/ c – p.1681

Multiply, multiply, partial fraction decomposition, I get c is equal to a × c - p + bp.1695

c is equal to ac - ap + bp.1706

c is equal to ac + -a + b × p.1715

Therefore, ac 1c, this coefficient is equal to the coefficient from here.1728

a is equal to 1, that takes care of the a.1737

I have got -a + p is equal to 0 because there is no p term here on the left,1741

which implies that a is equal to b which implies that b is also equal to 1.1750

Our c/ p × c - p is equal to actually 1/ p + 1/ c – p.1761

This is our a and this is our b.1771

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,1774

which is equal to the natlog of p - the natlog of c – p.1796

Therefore, what we have is c – p.1809

This is the left side of the integral.1820

Let us go ahead and rewrite what is it that we actually did.1834

We had the integral of c/ p × c - p = the integral of k dt.1843

We just did that integral, that is ln of p - ln of c - p = the integral of this is just kt + some constant.1853

I will go ahead write out constant, I do not want to use the word c1870

because I do not want you to confuse it with this c right here.1873

Over here, we have the natlog of p/ c - p is equal to kt + a constant.1878

I’m going to flip this and I’m going to write this as the natlog of c - p/ p = -kt - the constant.1895

I hope that make sense.1908

Log of a/b is log(a) – log(b).1910

If I flip this, it just takes the negative sign out here.1914

I move the negative sign to this side, now I have got that.1917

And now I exponentiate both sides.1922

I end up with c - p/ p is equal to e ⁻kt × e ⁻some constant.1926

This constant, I'm just going to call a.1942

I’m going to separate this out, this is going to be c/p - p/p which is 1 = a × e ⁻kt.1952

I have got, this is going to end up giving me c/p = a × e ⁻kt + 1.1968

When I solve this for p, I’m going to move the p up here, move this down here.1985

I'm left with p is equal to c which is the carrying capacity not the constant, = a which is some constant e ⁻kt + 1.1989

This is my solution to the logistic equation.2002

My population is equal to the carrying capacity divided by this thing,2007

some constant a × the exponential, some growth constant t + 1.2012

Let us go ahead and deal with the initial value.2019

Initial value, it is said that the population at time 0 is equal to some initial population.2023

Therefore, our p(0) is equal to c.2032

We use our equation ae ⁻kt × 0 + 1.2041

They tell me that it is equal to p(0), the initial population.2050

e⁰ is just 1, what I have here is c / a + 1 = p(0).2055

I rearrange this, c/ p(0) = a + 1.2068

I get a is equal to c/ p(0) – 1.2084

This is one version of it or when I do a common denominator, I get a = c – p(0)/ p0.2091

Our equation that we solved was the carrying capacity/ a × e ⁻kt + 1.2104

This value of a, I get it by taking the carrying capacity - the initial population divided by the initial population.2112

I have a way of finding a.2120

Let us go back to blue.2129

Our initial value problem dp dt = kp × 1 - the population/ c, where c is the carrying capacity.2133

p(0) = p sub 0, some number, which is the initial population.2165

It gives the following.2172

Our solution is, the carrying capacity divided by a × e ⁻kt + 1,2174

where a is equal to the carrying capacity - the initial population/ the initial population.2186

This is the solution to our problem.2193

What happens as t goes to infinity, as time just grows?2198

Let us analyze this solution, what happens when t goes to infinity?2204

When t goes to infinity, e ⁻kt goes to 0.2219

a × e ⁻kt goes to 0.2228

As this goes to 0, the population goes to c.2238

As time increases, the population goes to c, the carrying capacity, exactly what we said.2246

A population will grow and then it will level off at the carrying capacity.2251

Let us actually see what the solutions look like.2260

I have a couple of cases.2266

Case 1, what if the initial population is less than c?2270

Let us just pick some numbers, let us say our initial population of 100 and a carrying capacity of 2000.2278

2000 is the maximum population in this environment.2286

We have a formula for a, a is the carrying capacity - the initial population/ the initial population2289

which is 2000 - 100/ 100, that is going to equal 19.2295

Case 2, it is where the initial population is actually bigger than the carrying capacity.2304

Let us pick some numbers.2312

Let us do an initial population of 3000, where the carrying capacity is 2000.2314

We have a, a is equal to the carrying capacity - the initial population2321

divided by the initial population which is equal to 2000 - 3000/ 3000.2333

It is going to be equal to -1/3.2343

Our case 1, our equation, p is going to equal the carrying capacity 2000 divided by a which is 19 × e ⁻kt + 1.2348

In our case 2, our population is going to be the carrying capacity 2000 divided by a which is -1/3 e ⁻kt + 1.2367

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.2382

k is the nature of the growth.2390

The value of k is something that we deduce from data that we have collected.2393

Let us take a look at what these look like.2398

I have taken a particular case, a particular initial population, a particular carrying capacity.2399

A particular initial population and a particular carrying capacity.2404

Here is what they look like.2410

Initial population of 100 for different values of k, k(0.5), k(0.7), k(0.9).2420

They all start at an initial population, they start to grow, and then they level off at the carrying capacity.2431

The carrying capacity is 2000.2439

For different value of k, same thing.2442

The behavior is the same, it is the nature of the growth.2444

This definitely is a much better model for how populations actually behave.2447

They will eventually start to grow, that is resources start to deplete,2452

the environment can only support so much so they achieve a maximum carrying capacity.2456

This is for when the initial population is less than the carrying capacity.2462

The initial population is 100, carrying capacity was 2000.2469

There is going to be initial growth and it is going to the exponential growth, but then it is going to level off.2474

That is what is happening here.2479

For the case where the initial population is greater than the carrying capacity,2481

the initial population of 3000 with a carrying capacity of 2000, it is going to just decay.2485

For different values of k, that is all it is.2492

These different graphs are just different values of k, that has to do with the nature of the environment.2494

It is going to decline until it reaches the carrying capacity and it is going to level off.2500

That is what is happening here.2506

Let us go ahead and do an example.2508

Solve the following initial value problem then use the solution to find the time,2512

when the population reaches 75% of its carrying capacity.2518

In this particular case, dp dt = 0.065.2524

This is our k, they gave us the k, in this case.2530

P1 p/c, this is our carrying capacity.2533

This is our c, our carrying capacity.2539

The initial population is 120.2540

We have a lot of information here.2544

We have our k which is equal to 0.065.2546

We have an initial population equal to 120.2551

We have a carrying capacity of 1500.2556

We know what our solution looks like.2561

First of all, let us find what a is.2563

We know what the solution is already, we solve this differential equation.2568

It is equal to c/a × e ⁻kt + 1.2571

Let us go ahead and find what a is.2578

I know what a is, a is equal to carrying capacity - initial population/ initial population.2581

It is equal to 1500 which is the carrying capacity - 120 which is the initial population/ 120.2590

I get an a value of equal to 11.5.2598

This is the value that I put in there.2602

I have a k value of 0.065, I put that value in there.2605

My equation is p is equal to the carrying capacity which is 1500, and that goes here.2610

I have already found most of the parameters here.2618

1500 divided by 11.5 × e⁰.065 t + 1, that is my solution.2621

I want to find the time when the population reaches 75% of its carrying capacity.2640

75% of c is just equal to 0.75 × 1500 which is equal to 1125.2651

The equation that I’m going to solve is, when the population is 1125.2663

I want to solve for t.2668

It is 1500/ 11.5 × e⁰.065 t + 1.2671

I want to solve for t.2684

I’m going to do 1125 × 11.5 e⁰.065 t + 1 = 1500.2691

I get 12937.5 e⁰.065 t + 1125 = 15002707

e⁰.065 t is equal to 0.028986 subtract 1125 divided by the 012937.5.2721

I end up with this, take the natlog of both sides.2737

The natlog of this, the natlog of that, I end up with a final value of t = 54.5.2742

I do not know the units.2751

The units could be seconds, hours, minutes, days, years, whatever.2751

It is just 54.5 time units.2755

Let us see what this looks like.2759

This is what this looks like.2762

An initial population of 120, the carrying capacity of 1500, it is right there.2763

It starts to grow, at the t value of 54.5 which is here, I labeled it as 54.48,2774

I hit a population of 1125 which is 75% of my carrying capacity, which happens to be up here.2784

That is all that is going on here, I hope that made sense.2794

Let us see here, what else can I do?2802

One last thing to round things out.2807

There is one final concept that comes up in your problems, may or may not come up, but just in case it does.2811

One final concept that can come up.2820

We speak about the rate of change of population dp dt.2832

Sometimes we speak of the relative rate of change of population.2836

There is a difference, here is what they are.2840

Relative rate, you need to know how to turn that word problem into some symbolic form.2843

Relative rate, this is the topic that can come up now for population.2857

Growth rate is what we already know, that is just dp dt.2868

Growth rate, it is the rate of change of the population per unit change in time.2876

It is the derivative, dp dt.2880

Relative growth rate or relative rate of growth or relative rate, you might see it that way too.2882

Relative growth rate, it is equal to dp dt, the growth rate divided by the population at the time.2893

Or you can do it this way, 1/p dp dt.2907

The growth rate is dp dt, the normal derivative.2913

The relative growth rate is the normal derivative divided by whatever the population happens to be.2916

In other words, it is how fast the population is changing relative to how many there are in the population.2923

Relative just means divide by that number.2932

It is a ratio that you are actually coming up with.2936

In terms of equations, it is going to look like this.2942

For the standard model, we said that dp dt is equal kp.2945

The rate of growth is directly proportional to the population at the time.2962

Relative growth version is this.2968

Relative growth version is 1/p × dp dt is equal to k.2975

In words, this one is going to say, the relative growth rate of the population is a constant.2987

They are the same equation.2993

The only difference between this equation and this equation is I actually just divided by p and I brought it over here.2994

In words, it is a same equation, there is no difference.2999

The solution that we got for the standard model is going to be the same solution here.3005

We are still going to separate variables and integrate.3012

The only difference is it is going to be worded differently.3014

Here they are going to say the rate of change in the population is directly proportional to the population.3016

Here they are going to say the relative rate of change in the population is constant.3023

You are going to say the relative rate of change divided by the population is equal to a constant.3028

That is all that is happening here.3034

Growth rate, relative growth rate, just divided by whatever the dependent variable is.3037

It is the same equation just worded differently.3047

You have to be aware of that, just worded differently.3055

Thank you so much for joining us here at www.educator.com.3064

We will see you next time, bye.3066