Population Growth: US Carrying Capacity & Predictions

[ineedhelpnow]

i suck at these kind of problems
#1
Suppose a population grows according to a logistic model with initial population 1000 and carrying capacity 10000. if the population grows to 2500 after one year, what will the population be after another three years.#2
a. make a guess as to the carrying capacity for the US population. Use it and the face that the population was 250 million in 1990 to formulate a logistic model for the US population.
b. determine the value k in your model by using the fact that the population in 2000 was 275 million.
c. use your model to predict the US population in the years 2100 and 2200.
d. use your model to predict the year in which the US population will exceed 350 million.

 

[I like Serena]

Penny for your thoughts...

 

[ineedhelpnow]

Quarter for yours...:)

im seriously lost. i don't know how to do these.

 

[I like Serena]

Quarter for yours...:)

im seriously lost. i don't know how to do these.

Well, your textbook has to explain what a "logistic model" is, and what "carrying capacity" is.
Can you quote the definitions for those?

 

[ineedhelpnow]

A population often increases exponentially in its early stages but levels off eventually and approaches its carrying capacity because of limited resources. if P(t) is the size of the population at time t, we assume that $\frac{dP}{dt}=kP$ if P is small. This says that the growth rate is initially close to being proportional to size. in other words, the relative growth rate is almost constant when the population is small. but we also want to reflect the fact that the relative growth rate decreases as the population P increases and becomes negative if P ever exceeds its carrying capacity M, the maximum population that the environment is capable of sustaining in the long rune.

and it gives the equation:
$\frac{dP}{dt}=kP(1-\frac{P}{M})$

 

[I like Serena]

A population often increases exponentially in its early stages but levels off eventually and approaches its carrying capacity because of limited resources. if P(t) is the size of the population at time t, we assume that $\frac{dP}{dt}=kP$ if P is small. This says that the growth rate is initially close to being proportional to size. in other words, the relative growth rate is almost constant when the population is small. but we also want to reflect the fact that the relative growth rate decreases as the population P increases and becomes negative if P ever exceeds its carrying capacity M, the maximum population that the environment is capable of sustaining in the long rune.

and it gives the equation:
$\frac{dP}{dt}=kP(1-\frac{P}{M})$

That means that you can solve this in 2 ways:

• Solve the differential equation (not so easy but doable) and fill in the numbers.
• Use for instance Euler's method to approximate the population after 3 years.

What do you think would be intended?

 

[ineedhelpnow]

solve the differential equation?

 

[I like Serena]

solve the differential equation?

It goes for instance like this:

\begin{array}{lcl}
\frac{dP}{dt}&=&kP(1-\frac P M) \\

\frac{dP}{P(1-\frac PM)} &=& k\,dt \\

\frac{dP}{1-(\frac{2P}{M}-1)^2} &=& \frac {kM}4\,dt \\
\end{array}

After this, substitute $u=\frac{2P}{M}-1$ and use that:
$$\frac {d}{du}(\text{artanh}(u)) = \frac{1}{1-u^2}$$

After some more steps you should find that:
$$P = \frac{e^{kt}}{C+M e^{kt}}$$

Hmm.
Doable, but I suspect you are supposed to use Euler's method...

 

[ineedhelpnow]

the book only gives this:

$P(t)= \frac{M}{1+Ae^{-kt}}$
where $A=\frac{M-P_0}{P_0}$

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i have no idea how to use euler's method for such a problem

 

[ineedhelpnow]

$P(1)=\frac{M}{1+Ae^{kt}}$

$2500=\frac{10000}{1+9e^k}$

$2500(1+9e^k)=10000$

$1+9e^k=4$

$9e^k=3$

$e^k=\frac{1}{3}$

$k=\ln\left({\frac{1}{3}}\right)=-\ln\left({3}\right)$$P(4)=\frac{10000}{1+9e^{(-ln(3)*4)}}$

did i do it right or did i just embarrass myself by typing up a whole bunch on nonsense?

 

[MarkFL]

Before we plug-n-chug, let's look at a method for solving the logistic equation. Let's write it in the form:

\(\displaystyle \frac{dP}{dt}=k_1P(1-k_2P)\) where \(\displaystyle P(0)=P_0\)

Now, if we separate variables, switch dummy variables and use the boundaries, we obtain:

\(\displaystyle \int_{P_0}^{P(t)}\frac{du}{u(1-k_2u)}=k_1\int_0^t\,dv\)

On the left, we can use partial fractions. we will assume the left integrand may be written as:

\(\displaystyle \frac{1}{u(1-k_2u)}=\frac{A}{u}+\frac{B}{1-k_2u}\)

Using the Heaviside cover-up method, we find:

\(\displaystyle A=1\)

\(\displaystyle B=k_2\)

Hence, our integral is now:

\(\displaystyle \int_{P_0}^{P(t)}\frac{1}{u}-\frac{k_2}{k_2u-1}\,du=k_1\int_0^t\,dv\)

Integrating, there results:

\(\displaystyle \left[\ln\left|\frac{u}{k_2u-1}\right|\right]_{P_0}^{P(t)}=k_1[v]_0^t\)

\(\displaystyle \ln\left|\frac{P\left(k_2P_0-1\right)}{P_0\left(k_2P-1\right)}\right|=k_1t\)

Convert from logarithmic to exponential form:

\(\displaystyle \frac{P\left(k_2P_0-1\right)}{P_0\left(k_2P-1\right)}=e^{k_1t}\)

Solving for $P$, we find:

\(\displaystyle P\left(k_2P_0-1\right)=e^{k_1t}\left(P_0\left(k_2P-1\right)\right)\)

\(\displaystyle P\left(k_2P_0-1\right)=e^{k_1t}\left(k_2P_0P-P_0\right)\)

\(\displaystyle P\left(k_2P_0-1\right)=k_2P_0Pe^{k_1t}-P_0e^{k_1t}\)

\(\displaystyle k_2P_0Pe^{k_1t}-P\left(k_2P_0-1\right)=P_0e^{k_1t}\)

\(\displaystyle P\left(k_2P_0e^{k_1t}-k_2P_0+1\right)=P_0e^{k_1t}\)

\(\displaystyle P=\frac{P_0e^{k_1t}}{k_2P_0e^{k_1t}-k_2P_0+1}\)

Now, with \(\displaystyle k_2=\frac{1}{M}\) and $k_1=k$, we may write:

\(\displaystyle P=\frac{MP_0e^{kt}}{P_0e^{kt}-P_0+M}\)

\(\displaystyle P=\frac{M}{1-e^{-kt}+\frac{M}{P_0}e^{-kt}}\)

\(\displaystyle P=\frac{M}{1+\left(\frac{M}{P_0}-1\right)e^{-kt}}\)

\(\displaystyle P=\frac{M}{\left(\frac{M-P_0}{P_0}\right)e^{-kt}+1}\)

If we define:

\(\displaystyle A=\frac{M-P_0}{P_0}\), the we may write:

\(\displaystyle P(t)=\frac{M}{Ae^{-kt}+1}\)

Now we "own" that formula. To answer the first question, we use the given data:

\(\displaystyle M=10000,\,A=\frac{10000-1000}{1000}=9\)

Hence, the general solution to this problem is:

\(\displaystyle P(t)=\frac{10000}{9e^{-kt}+1}\)

To determine $k$, we may use the fact that:

\(\displaystyle P(1)=2500\)

\(\displaystyle \frac{10000}{9e^{-k}+1}=2500\)

\(\displaystyle \frac{4}{9e^{-k}+1}=1\)

\(\displaystyle 4=9e^{-k}+1\)

\(\displaystyle 3=9e^{-k}\)

\(\displaystyle 3=e^{k}\)

\(\displaystyle k=\ln(3)\)

And so we have the solution:

\(\displaystyle P(t)=\frac{10000}{9e^{-t\ln(3)}+1}\)

\(\displaystyle P(t)=\frac{10000}{9e^{\ln\left(3^{-t}\right)}+1}\)

\(\displaystyle P(t)=\frac{10000}{9\cdot3^{-t}+1}\)

\(\displaystyle P(t)=\frac{10000}{3^{2-t}+1}\)

After another 3 years, we have $t=4$, hence:

\(\displaystyle P(4)=\frac{10000}{3^{2-4}+1}=\frac{10000}{3^{-2}+1}=\frac{10000}{\frac{10}{9}}=9000\)

This is equivalent to the answer you give, but as you can see, we can use the properties of exponents and logarithms to greatly simplify. :D

 

[ineedhelpnow]

THANK YOU! yeah my book does the whole thing were it goes through a bunch of shtuff until it gets to the final equation, which then you plug in the numbers. how do i do number 2 though?

 

[MarkFL]

THANK YOU! yeah my book does the whole thing were it goes through a bunch of shtuff until it gets to the final equation, which then you plug in the numbers. how do i do number 2 though?

Well, let's begin with part a). What would you say is the carrying capacity of the U.S. population?

 

[ineedhelpnow]

not 250 million?

 

[MarkFL]

not 250 million?

That's the given population for 1990, and we have increased since then, so our carrying population must be greater. Consider that there are over a billion people in India, and India is smaller than the U.S. in terms of area. What kind of guess would you say is the number of people that the U.S. can sustain, you really only need to make a guess, and I would use a fairly round number too, perhaps rounded to the nearest 100 million.

 

[ineedhelpnow]

that's what I am confused about because if that's the case can't i pick ANY large number.

 

[MarkFL]

that's what I am confused about because if that's the case can't i pick ANY large number.

Well, you could, but I would make an effort to make a good educated guess, so that your model is reasonable. You probably would not want to say 100 billion, as that seems unreasonable to me. What number seems plausible to you?

 

[ineedhelpnow]

400 million or 500 million

 

[MarkFL]

400 million or 500 million

Given that India can sustain over a billion people, why would you think we could sustain less than half of that? If you have reasons for this, that's fine, but in my opinion we could sustain well over a billion people, perhaps as many as 2 billlion...or even more.

 

[ineedhelpnow]

oh my bad. I am thinking of how many people probably live in the us. now how much the us can carry. how about ~3 billion.

 

[MarkFL]

oh my bad. I am thinking of how many people probably live in the us. now how much the us can carry. how about ~3 billion.

Okay, I don't think that's unreasonable. So, how can you use this, and the fact that in 1990 the U.S. population was 250 million? When should you set time $t=0$ and what unit do you think is appropriate for the population so you aren't dealing with huge numbers?

 

[ineedhelpnow]

i really don't know. i have like an idea sort of, of what the equation would look like but it's not complete.

 

[MarkFL]

Well, we know the equation is going to look like:

\(\displaystyle P(t)=\frac{M}{Ae^{-kt}+1}\)

And we have decided that $M=3\E{9}$. Now I would choose a unit of 100 million people, so that $M=30$. I would also let time $t=0$ be the year 1990 as instructed. Hence:

$P_0=2.5$

So, what is $A$, and using this, what is your population curve?

 

[ineedhelpnow]

$A=\frac{M-P_0}{P_0}$

so $A=\frac{30-2.5}{2.5}=11$

is that right?

 

[MarkFL]

$A=\frac{M-P_0}{P_0}$

so $A=\frac{30-2.5}{2.5}=11$

is that right?

Yes, good! :D So we have:

\(\displaystyle P(t)=\frac{30}{11e^{-kt}+1}\)

So, now you want to determine $k$, using the information from part b):

\(\displaystyle P(10)=2.75\)

And then you can answer the rest of the problem. (Yes)

 

[ineedhelpnow]

alright. thanks. give me 5 days to solve it and then ill come back with an answer. got to get started! bye :)

 

[ineedhelpnow]

for part c) do i set my t as 20 and 30?

 

[MarkFL]

for part c) do i set my t as 20 and 30?

No, since 1990 corresponds to $t=0$, then 2100 would be $t=2100-1990=110$ and likewise 2200 would be $t=210$. :D

 

[ineedhelpnow]

haha i was thinking 2010 and 2020

 

[MarkFL]

haha i was thinking 2010 and 2020

You would have been correct in that case. :D

 

[ineedhelpnow]

b)
$2.75(11e^{-10k}+1)=30$
$11e^{-10k}+1=10.91$
$11e^{-10k}=9.91$
$e^(-10k}=0.90083$
$-10k=ln(0.90083)$
$k=0.010444$

c)
$P(110)=\frac{30}{11e^{(-0.010444)*110}+1}$
put that into wolframalpha and you get 6.69 billion

$P(210)=\frac{30}{11e^{(-0.010444)*210}+1}$
put that into wolframalpha and you get 13.5 billion

d)
$3.5=\frac{30}{11e^{-0.01044t}+1}$
solve that for t and you get 24 so 2014

did i do them right?

 

[MarkFL]

b) Let's use:

\(\displaystyle 11e^{-10k}+1=\frac{120}{11}\)

\(\displaystyle 11e^{-10k}=\frac{109}{11}\)

\(\displaystyle e^{-10k}=\frac{109}{121}\)

\(\displaystyle e^{-k}=\left(\frac{109}{121}\right)^{\frac{1}{10}}\)

Now substituting into the population function, we obtain:

\(\displaystyle P(t)=\frac{30}{11\left(\frac{109}{121}\right)^{\frac{t}{10}}+1}\)

c) Now, in the year 2100, when $t=110$, we find:

\(\displaystyle P(110)=\frac{30}{11\left(\frac{109}{121}\right)^{11}+1}\approx6.67\)

Now, recall that we chose as units, 100 million people, and so this equates to about 667 million people. We should suspect something is wrong if our answer exceeds the carrying capacity of 3 billion.

And, in the year 2200, when $t=210$, we find:

\(\displaystyle P(210)=\frac{30}{11\left(\frac{109}{121}\right)^{21}+1}\approx13.47\)

This equates to about 1.347 billion people.

d) Let $P(t)=3.5$ and solve for $t$:

\(\displaystyle 3.5=\frac{30}{11\left(\frac{109}{121}\right)^{\frac{t}{10}}+1}\)

\(\displaystyle 3.5\left(11\left(\frac{109}{121}\right)^{\frac{t}{10}}+1\right)=30\)

\(\displaystyle 11\left(\frac{109}{121}\right)^{\frac{t}{10}}=\frac{53}{7}\)

\(\displaystyle \left(\frac{109}{121}\right)^{\frac{t}{10}}=\frac{53}{77}\)

\(\displaystyle t=10\frac{\ln\left(\frac{53}{77}\right)}{\ln\left(\frac{109}{121}\right)}\approx36\)

Thus, the year is 2026

 

[ineedhelpnow]

so solving for k doesn't work?

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and where did 120/11 come from?

 

[MarkFL]

so solving for k doesn't work?

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and where did 120/11 come from?

Solving for $k$ is fine, but recall the problem to which you linked...I had kind of forgotten that we can just solve for $e^{-k}$ instead of having to go through using the rules of exponents/logs just to get back to where we would be if we just use $e^{-k}$.

The 120/11 is the exact value of the approximation 10.91 that you used, by dividing 30 by 2.75:

\(\displaystyle \frac{30}{2.75}=\frac{3000}{275}=\frac{25\cdot120}{25\cdot11}=\frac{120}{11}\)