For more information, please see full course syllabus of AP Calculus AB

For more information, please see full course syllabus of AP Calculus AB

### Population Growth: The Standard & Logistic Equations

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

- Intro 0:00
- Standard Growth Model 0:30
- Definition of the Standard/Natural Growth Model
- Initial Conditions
- The General Solution
- Example I: Standard Growth Model 10:45
- Logistic Growth Model 18:33
- Logistic Growth Model
- Solving the Initial Value Problem
- What Happens When t → ∞
- Example II: Solve the Following g Initial Value Problem 41:50
- Relative Growth Rate 46:56
- Relative Growth Rate
- Relative Growth Rate Version for the Standard model

### AP Calculus AB Online Prep Course

### Transcription: 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 equation *0082

*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 population*0116

*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 data *0254

*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 model*1172

*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 equation*1309

*but you can extract a lot of information from just looking at the equation*1312

*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 - p *1645

*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 c*1870

*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 population*2289

*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 population*2321

*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 = 1500*2707

*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

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