Logistical Growth model

In summary, the problem involves finding the rate of change in the population of giraffes on an island using a logistic growth model. This requires writing a function P(t) that models the population t years after 1970 and taking its derivative. The formula for P(t) is M/(1+ Ae^-kt) where M is the maximum capacity and A is a constant. The derivative of this function is (kM*e^-kt)/(1+ Ae^-kt)^2. Plugging in the values for M and A, and solving for k, the rate of change in the population can be calculated at a certain time.
  • #1
CaptainOfSmug
13
0

Homework Statement


The growth of a giraffe population on an island follows a logistical growth model. The maxium giraffe population on the island is estimated at 1000. The population was first measured in 1970 at 125. Then in 1975 the population was 158

a) Find the rate of change in the population in 1990 by first writing a function P(t) that models the population t years after 1970, then taking the derivative of that function

b.) Find the rate of change in the population in 1990 by using the logistic differential equation [/B]

Homework Equations


P(t)= M/ 1+ Ae-kt Where M is the maximum capactiy and
A= (M-P0)/(P0)
Also: dP/dt=kP(1-(M/P))

The Attempt at a Solution


Well part A is what is really stumping me, I'm not completely sure how to write a function for this... and it's making me feel silly. I've been scouring my textbook and they have know written examples, or exercises to give me a hint. My first thought was to just find the slope given the two points, but the more I thought about it it doesn't make sense because that would indicate exponential growth, which isn't the case for the problem. I need some hints, about that part, I'm sure i can derive it by myself.

For part b I think I did it right, but maybe it's not really the rate of change. I did as follows, with P0=125 and P5=158:
P(t)=1000/(1+7e-kt the 7 coming from the second equation I wrote to solve for "A"
I then used that fact that in 1975 the population was 158 so:
158=1000/(1+7e-k5)
k=(ln(421/553))/-5
k is approximately 298.38

To me this seems like a reasonable population size in 1990, but I don't think it necessarily reflects a rate of change. I know I am missing some important part but I can't figure it out so far

Any help is appreciated! Thanks in advance!
 
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  • #2
CaptainOfSmug said:

Homework Statement


The growth of a giraffe population on an island follows a logistical growth model. The maxium giraffe population on the island is estimated at 1000. The population was first measured in 1970 at 125. Then in 1975 the population was 158

a) Find the rate of change in the population in 1990 by first writing a function P(t) that models the population t years after 1970, then taking the derivative of that function

b.) Find the rate of change in the population in 1990 by using the logistic differential equation [/B]

Homework Equations


P(t)= M/ 1+ Ae-kt Where M is the maximum capactiy and
A= (M-P0)/(P0)
Also: dP/dt=kP(1-(M/P))

The Attempt at a Solution


Well part A is what is really stumping me, I'm not completely sure how to write a function for this... and it's making me feel silly. I've been scouring my textbook and they have know written examples, or exercises to give me a hint. My first thought was to just find the slope given the two points, but the more I thought about it it doesn't make sense because that would indicate exponential growth, which isn't the case for the problem. I need some hints, about that part, I'm sure i can derive it by myself.

For part b I think I did it right, but maybe it's not really the rate of change. I did as follows, with P0=125 and P5=158:
P(t)=1000/(1+7e-kt the 7 coming from the second equation I wrote to solve for "A"
I then used that fact that in 1975 the population was 158 so:
158=1000/(1+7e-k5)
k=(ln(421/553))/-5
k is approximately 298.38

To me this seems like a reasonable population size in 1990, but I don't think it necessarily reflects a rate of change. I know I am missing some important part but I can't figure it out so far

Any help is appreciated! Thanks in advance!

You write
[tex] P(t) =\frac{M}{1} + Ae^{-kt} = M + A e^{-kt}[/tex]
Did you mean
[tex] P(t) = \frac{M}{1 + A e^{-kt}}?[/tex]
If so use parentheses, like this: M/(1+e^(-kt)) or this: M/(1 + A e-kt).

Anyway, are you saying you do not know how to find the derivative of P(t)? Well, what rules do you know about derivatives? What kinds of derivatives have you done so far?
 
  • #3
Sorry I forgot the parenthesis on that one yes, it is M/(1+Ae-kt)
I guess I'm not sure exactly what the question is asking, do I just use the above formula and derive it to get the rate of change? Would I have to find a value of k first or just leave it as a constant?
 
  • #4
CaptainOfSmug said:
Sorry I forgot the parenthesis on that one yes, it is M/(1+Ae-kt)
I guess I'm not sure exactly what the question is asking, do I just use the above formula and derive it to get the rate of change? Would I have to find a value of k first or just leave it as a constant?

The parameter k IS just a constant, so does not change the nature of the derivative---only some of the details in the final formula. Of course you need to plug in a numerical value for k eventually, but it will not make any difference when you do that, provided that you do not perform excess numerical roundoff during the calculations. My own personal preference is to keep the symbolic constant throughout, and just substitute a number at the last minute---because you often (not always) get more "revealing" formulas that way. That is, you can often better see the final effects of different factors on the answer.
 
  • #5
Okay so deriving that function I got:
P'(t)= (7000e-kt)/(1+7e-kt)^2
I found my value of k to be (1/5)ln(421/553)
Plugging in that value and letting t=20 years the answer was 4.3*10-4

I'm assuming part a) and part b) should be equivalent. Can you give me any advice on what I'm doing wrong? I could be wrong but I thought the k value was the rate of change in these problems?
 
  • #6
CaptainOfSmug said:

Homework Statement


The growth of a giraffe population on an island follows a logistical growth model. The maxium giraffe population on the island is estimated at 1000. The population was first measured in 1970 at 125. Then in 1975 the population was 158

a) Find the rate of change in the population in 1990 by first writing a function P(t) that models the population t years after 1970, then taking the derivative of that function

b.) Find the rate of change in the population in 1990 by using the logistic differential equation [/B]

Homework Equations


P(t)= M/ 1+ Ae-kt Where M is the maximum capactiy and
A= (M-P0)/(P0)
Also: dP/dt=kP(1-(M/P))

Should this not be [tex]
\frac{dP}{dt} = kP\left(1 - \frac{P}{M}\right)[/tex]

[...]

For part b I think I did it right, but maybe it's not really the rate of change. I did as follows, with P0=125 and P5=158:
P(t)=1000/(1+7e-kt the 7 coming from the second equation I wrote to solve for "A"
I then used that fact that in 1975 the population was 158 so:
158=1000/(1+7e-k5)
k=(ln(421/553))/-5
k is approximately 298.38

To me this seems like a reasonable population size in 1990, but I don't think it necessarily reflects a rate of change. I know I am missing some important part but I can't figure it out so far

Part (a) asks you to find [itex]P'(20)[/itex] by differentiating [itex]P[/itex].

Part (b) asks you to find [itex]P'(20)[/itex] by using the fact that [itex]P[/itex] satisfies the logistic growth ODE:
[tex]P'(t) = kP(t)\left(1 - \frac{P(t)}{M}\right).[/tex] The two answers should, of course, be equal.
 
  • #7
Yes my mistake, the equation should be as you stated. I guess I'm confused about what function to use to differentiate, am I supposed to solve the differential equation
dP/dt=kP(1-(P/M)?, I'm assuming I would use the k value I found in my first post. In that case I suppose I would just integrate both sides of the equation, find the "c" constant, plug it back in and use my k value and then solve for P? However, once P is solved for and I take the derivative it would just be the integral I started with wouldn't it? This problem is giving me a complete headache haha!
 
  • #8
CaptainOfSmug said:
Yes my mistake, the equation should be as you stated. I guess I'm confused about what function to use to differentiate, am I supposed to solve the differential equation
dP/dt=kP(1-(P/M)?, I'm assuming I would use the k value I found in my first post. In that case I suppose I would just integrate both sides of the equation, find the "c" constant, plug it back in and use my k value and then solve for P? However, once P is solved for and I take the derivative it would just be the integral I started with wouldn't it? This problem is giving me a complete headache haha!

You were already given the solution to the DE, but for practice, you should plug it back into the DE to verify that fact. You were given enough information in the problem to determine all the parameters, but you made some errors in your Post #1: when I (or, rather, Maple) did it I got a value of ##k## between 0.01 and 0.10; I will not tell you the exact value, since that would spoil your fun of getting the answer for yourself. Obviously, your ##k## value of 298.38 is very different from mine.
 
  • #9
Okay so I got my k value to be (-1/5)ln(421/553) which is approximately 0.0545. The 298.38 came from the fact of using that value in P(20), which just gives me the population of giraffes in 1990... which is not what the question is asking I now realize *facepalm*
but plugging that k value into the DE I get:
P(t)=e0.054490489t+125 I should note that I found my constant "c" to be ln(125)

I feel like I am right on the verge of this thing, my main problem is how the questions are written I suppose.
For part a) am I supposed to find k, and then put it into the equation dP/dt=kP(1-(P/M)) separate the equation, integrate, find the constant, and then come up with the function P(t)? When I did that I got the formula P(t)=e.054490489t+125. However, the instructions state that I am supposed to derive that, when I did that I did not get the logistic differential equation??
or if I use my answer for what P(t) equals, and set it to the logistic equation of (1000)/(1+7e-kt) my answers are not equivalent. For example, I let t=0 for both sides of the equation which came out to 126=125

For part b) am I supposed to use the derived formula from about, and then solve for P(20)?

Thanks for being patient with me, I just really want to understand this problem!
 
Last edited:
  • #10
CaptainOfSmug said:
Okay so I got my k value to be (-1/5)ln(421/553) which is approximately 0.0545. The 298.38 came from the fact of using that value in P(20), which just gives me the population of giraffes in 1990... which is not what the question is asking I now realize *facepalm*
but plugging that k value into the DE I get:
P(t)=e0.054490489t+125 I should note that I found my constant "c" to be ln(125)

I feel like I am right on the verge of this thing, my main problem is how the questions are written I suppose.
For part a) am I supposed to find k, and then put it into the equation dP/dt=kP(1-(P/M)) separate the equation, integrate, find the constant, and then come up with the function P(t)? When I did that I got the formula P(t)=e.054490489t+125. However, the instructions state that I am supposed to derive that, when I did that I did not get the logistic differential equation??

For part b) am I supposed to use the derived formula from about, and then solve for P(20)?

Thanks for being patient with me, I just really want to understand this problem!

I read part (b) as asking for ##dP(t)/dt## at ##t = 20##. Look again at the words, and you will see that is what they say (after you understand that "rate of change" = ##dP/dt##).
 
  • #11
Okay now I am lost... :( I'm assuming I did part a incorrectly then because when P'(t)=e-.05449048t letting t =20 I get a population of 127.97... doesn't make sense.
When I use my k value for P(20)=(1000)/(1+7e-.054545034(20)) I get a population of 298.38 which seems much more reasonable.

EDIT*

Okay I finally figured out the answer, it didn't look pretty like the logistic model but it definitely was equal. Thanks for the guidance and for putting up with my poorly worded questions :P
Cheers!
 
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1. What is the Logistical Growth model?

The Logistical Growth model is a mathematical model used to describe the growth of a population over time. It takes into account the maximum possible population size, the initial population size, and the rate at which the population is growing.

2. How is the Logistical Growth model different from the Exponential Growth model?

The main difference between the two models is that the Exponential Growth model assumes unlimited resources for growth, while the Logistical Growth model takes into account the carrying capacity of the environment. This means that the Logistical Growth model predicts a slower and more realistic growth rate for a population.

3. What is the importance of the carrying capacity in the Logistical Growth model?

The carrying capacity, or the maximum sustainable population size of an environment, is a crucial factor in the Logistical Growth model. It allows us to predict when a population will reach its maximum size and when it will start to slow down in its growth rate. It also helps us understand how changes in the environment can affect the growth of a population.

4. How do scientists use the Logistical Growth model in their research?

The Logistical Growth model is commonly used in ecology and population biology to study the growth and dynamics of populations in different environments. It is also used in other fields, such as economics and epidemiology, to understand how populations may change over time and how to manage their growth.

5. Are there any limitations to the Logistical Growth model?

Like any mathematical model, the Logistical Growth model has its limitations. It is based on certain assumptions, such as a constant growth rate and a stable environment, which may not always hold true in real-life situations. Additionally, it may not accurately predict sudden changes or fluctuations in a population. Therefore, it is important for scientists to use caution when applying this model to real-world scenarios.

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