Main content

## AP®︎/College Calculus BC

### Course: AP®︎/College Calculus BC > Unit 7

Lesson 9: Logistic models with differential equations# Worked example: Logistic model word problem

Finding the carrying capacity of a population that grows logistically. Also finding the population's size when it's growing the fastest.

## Want to join the conversation?

- In a previous video Sal stated the formula

dn/dt = Rn (1 - N/K)

in the above video he then used the formula

Nr(K-N).

How are these interchangeable?

Many thanks.(15 votes)- Proof that dN/dt = Nr * (1 - N/K) is the same as dN/dt = Nc * (K-N), where c = r/K:

Start with:

dN/dt = Nr * (1 - N/K)

We know that (1 - N/K) = 1/K * (K-N)

dN/dt = Nr * 1/K (K - N)

dN/dt = Nr/K * (K-N)

r and K (carrying capacity) are just constants. So, let's make r/K = c, a constant.

dN/dt = Nc * (K-N)

Now, we've shown that the two forms are equivalent.

But note that in the second, we have Nc * (K-N), while in the first we have Nr * (1 - N/K). c is NOT equal to r. In fact, c = r/K.

Sal Khan used dP/dt = kP (a - P) which is the same as the second form, dN/dt = Nc * (K-N). Here, P is N, a is uppercase K, and lowercase k is c.

Ask if you have questions! And let me know if I made any errors. Thanks!(26 votes)

- At07:10. The population's size is the half-way between the initial population(P) and maximum population(a). The initial population is 700, not 0. Shouldn't the middle point be shifted by 350?(9 votes)
- The initial population is 700, but this is where t=0. What Sal did was finding the vertex of dP/dt, which is a function of P, not t. The vertex is halfway between the points where dP/dt is equal to 0. dP/dt is equal to 0 when P is equal to 0. This is not the same thing as the initial population. Therefore the middle point is not shifted over.(9 votes)

- Is it possible to find the fastest growth by finding the derivative of the logistic equation, and then locating the inflection point?(4 votes)
- Yes. The fastest growth would occur when the derivative is maximized. To maximize the derivative, we find where
*it's*derivative is 0, i.e. where the second derivative is 0, i.e. an inflection point.(8 votes)

- We would all appreciate it if Sal could explain why he uses an equation that is completely different from the ones he teaches in the surrounding videos.

The cognitive dissonance is real. It's like we're driving down the road all happy and learning... and up pops a brick wall out of nowhere and we just smash into it(7 votes) - At6:33, when Sal says dpt, shouldn't it say dp/dt?(6 votes)
- yes it should be dp/dt, the rate of change of population with respect to time(2 votes)

- 6:50How do you know the max growth is halfway between zeros ?(2 votes)
- 𝑑𝑃∕𝑑𝑡 is a quadratic function (when graphed over 𝑃), which is symmetric about the vertex.(9 votes)

- So will the population ever reach the max capacity or will it just get very close to the max?(3 votes)
- Can we take second derivative to find the max?

Let y = dP/dt = P x 48000/4000 - P^2/4000

y'= 48000/4000 - 2P/4000 = 0

2P/4000 = 48000/4000

2P = 48000

P = 24000(2 votes)- That works too! Understand why it does though.

When you find the derivative of $\frac{dP}{dt}$, what you're doing is finding critical points of the function. Now, when you equate it to zero and find the value of P, it is the value for which $\frac{dP}{dt}$ is a maximum or minimum. You can do a second derivative test (or, as our function itself is a derivative, this would technically be a third derivative test) to see if the value of P is a maximum or minimum. As it should come to be a maximum, P = 24,000 would be the maximum of the function $\frac{dP}{dt}$. So, at P = 24,000, $\frac{dP}{dt}$ is maximum, which is exactly what we needed to show!(2 votes)

- why does Sal introduce the logistic differential equation in the form dP/dt=kP(1-P/a) in the previous videos, but here he switches to the form dP/dt=kP(a-P) ?(2 votes)
- I assume he simplified $kP(1-\frac{P}{a})$ to $kP(\frac{a-P}{a})$ and then took $\frac{k}{a}$ = k. Though yes, I do agree that he should've touched upon that a bit before using it(1 vote)

- at first "a" was the carrying capacity, which makes sense. Then, why did he find "p" and call it the carrying capacity again??(2 votes)
- Where exactly does he say that? He seems to refer to "a" as the carrying capacity the whole time.(1 vote)

## Video transcript

- [Narrator] The population
P of T of bacteria in a petry dish satisfies the logistic differential equation. The rate of change of
population with respect to time is equal to two times the population times the difference between six and the population divided by 8000, where T is measured in hours
and the initial population is 700 bacteria. What is the carrying
capacity of the population; and what is the population's size when it's growing the fastest? So in order to even attempt
to answer these questions at any point if you're inspired,
definitely pause the video and try to answer it by yourself. Let's just remind ourselves
what we're talking about or what they're talking about with the logistic differential equation and the carrying capacity. So in general a logistic
differential equation is one where we seeing
the rate of change of, and it's often referring to population, so let's just stick with population. So the rate of change of our
population with respect to time is proportional to the
product of the population and the difference between what's known as the carrying capacity
and the population. Now why is this a model
that you'll see a lot, especially why is it useful for studying things like populations. When a population is small the environment really isn't limiting it and so assuming it starts
from some none zero value, this thing grows, this
thing is not going to get much smaller and so
our population is going to a rate of change is going to increase. And so let me just draw
a little graph here to show the typical solution to a logistic differential equation. So this is our population, this is time, so when our population is low, let's say it's gonna start
from some non zero value, if it was zero what would happen, well then our rate of
change would just be zero and our population would never
grow and that makes sense. If you have no bunnies on your island then there never will be
any bunnies on your island but if you have a few
bunnies well initially, their rate of change, the rate of population is
gonna keep accelerating as this thing grows, it's
gonna keep accelerating but then at some point your
environment is going to limit how many bunnies
for example or bacteria can grow in your environment. Because once the
population gets close to A, this thing right over here
is going to approach zero and its going to make our rate of change smaller and smaller and smaller. And so you can imagine
in the limiting case as P gets very close to A, you can imagine as T approaches infinity, our rate of change is
going to approach zero. So one way to think about it is, our population would acemtote
towards the carrying capacity. That is A right over here,
that is our carrying capacity. So there's a couple of ways of answering this first question; one way is we can actually put our
logistic differential equation in this form and then we can recognize what the carrying capacity is. The other way is to think
about, well what happens as T approaches infinity. As T approaches infinity,
this thing approaches zero and so we can think from this
logistic differential equation what P values would
make this thing be zero based on this differential equation or when this thing approaches zero, what P values would this approach. So let's do it both ways. So one way, let me write it
in this form right over here, so it's close, this is
six minus P over 8000, can we make it in the form
where we just have some number minus P. Well if we multiply this times 8000 and then we divided this by 8000 we wouldn't be changing the
value of that expression so let's do that. If we divide this by 8000 you have DP DT is equal to 2P over
8000 is P over 4000 times and now let's multiply
this expression by 8000. So six times 8000 is 48,000
minus P over 8000 times 8000 is minus P and there you have it. We've written it down
to somewhat classic form and we can see that the carrying capacity is 48,000, I guess we say 48,000 bacteria. So this thing right over here is 48,000. Another way we could think about it is, well the carrying capacity is what happens as T approaches infinity. So as T approaches infinity,
this thing right over here approaches zero, you can
see the rate of change approaches zero. So when that approaches
zero what does P approach? So we can just solve for
this; six minus P over 8000. Well there's two situations, two Ps for which our derivative is equal to zero. There's one case when
this is equal to zero, in which case our population is to zero or the other case is when
this is equal to zero. So let's do that; six minus
P over 8000 is equal to zero or we could say P over
8000 is equal to six and so we can say multiply
both sides by 8000, P is equal to 48000. Which is exactly what
we had right over there. So now let's answer the second part. What is the population's size
when it's growing the fastest? So intuitively you can see
when that is right over here, the rate of change is gonna grow but then as we approach
the carrying capacity, the rate of change is
gonna start slowing down. So your maximum rate of change,
it's growing the fastest, is right about there but how
do we figure out it exactly. Well you could go back to the logistic differential equation. You can see that it's
really our rate of change is a function, you could view it as a function of our
population right over here. And this is actually a
quadratic expression, this is a concave downward
quadratic expression, it would look like this, if you were graphing rate
of change, let me do that, if you were to graph rate of change, DP DT as a function of population, well when the population is
small, so when the population is, say 700 or that's
where we're starting. I'll just speak in generalities, when your population is
small DPT, your rate is small but then it increases and
then some point around there the rate of change starts to
decrease and it approaches zero as our population is approaching
the carrying capacity. So this right over here for
example, would actually be our carrying capacity. What is this maximum point right over here and there's a couple of
ways you could approach it with calculus and even
algebra, we have many tools for identifying this maximum point which is really just a vertex
of this downward opening, this concave downward parabola. So let's just do that and
let's find this vertex and the vertex is just
half way between the zeros of this quadratic. So let's find the P values
that make this equal to zero which we actually just figured out, the vertex is just gonna
be half way between that. So when the population is zero
our rate of change is zero when the population is A,
which we know is 48,000, our rate of change is zero and
so our maximum rate of change is gonna happen half way
between those two points which is 24,000 So it is a population of 24,000. When we got that by really just saying well at one point does this quadratic, quadratic is a function of P, when does that hit a maximum point. Well that's going to be
half way between the zeros, the zeros happen when P equals
zero and P is equal to 48,000 so that's gonna happen when
the population is 24,000. So this type of problem seems
very intimidating at first, logistic differential equation, how do I actually solve
this and then analyze it but the key is to; one, recognize a logistic
differential equation, see what it's talking about and then maybe think of this
in terms of our rate of change as a function of our population. And remember that the carrying
capacity is what happens as T approaches infinity
and as T approaches infinity our rate of change approaches zero. And so if that's approaching zero, what must our population be approaching.