### Video Transcript

In this video, we will introduce
the logistic model for representing population growth. We’ll first recap the model used
for simple population growth and discuss why this model may not be appropriate in
certain circumstances. We’ll then look at how we can
develop this model further to take account of some of the practical constraints of
population growth, such as the maximum population that can be supported. And we’ll then consider a series of
examples in which we work with this new model.

You should already be familiar with
a model for simple population growth, in which the population increases at a rate
proportional to the population itself. Such population growth can be
modelled by the differential equation d𝑃 by d𝑡 equals 𝑘𝑃, where 𝑘 is the
constant of proportionality or the rate at which the population is growing. By separating the variables in this
differential equation and then integrating, we know that the general solution to
this differential equation is of the form 𝑃 equals 𝐴𝑒 to the power of 𝑘𝑡. And if we know the initial value of
the population at time 𝑡 equals zero, 𝑃 nought, we can find the particular
solution to be 𝑃 equals 𝑃 nought 𝑒 to the power of 𝑘𝑡. So we find that the population
grows exponentially. And so the graph of the population
over time would look a little like this.

However, we need to consider the
practicality of this model because it implies that the population will continue to
grow at an ever-increasing rate. In reality, this is unlikely to be
possible as the environment in which the population live will not have unlimited
resources. There will perhaps come a point
where the population is too large to be supported by the available resources. And at this point, our model will
cease to be accurate in predicting how this population will change over time. We therefore want to change our
model to include two new assumptions.

Firstly, the relative growth of the
population will decrease as the population itself increases. So whilst exponential growth might
be appropriate initially, the rate at which the population grows will eventually
slow down. The second assumption is that there
is a maximum population that can be supported in a given environment. This is known as the carrying
capacity. And we denote it using the letter
𝐿.

We’ll look at the model that
incorporates these assumptions in a moment. But the graph of the solution to
this model would look a little something like this. The population initially grows in
an exponential way. But this rate of growth then
decreases. And finally, the population levels
off at a level that is equal to the carrying capacity, 𝐿, of the environment.

The simplest model for population
growth, which incorporates both of the assumptions we just discussed, is as
follows. One over 𝑃 d𝑃 by d𝑡 equals 𝑘
multiplied by one minus 𝑃 over 𝐿. And proving this is beyond the
scope of this video. Multiplying both sides of this
equation by 𝑃 gives d𝑃 by d𝑡 equals 𝑘𝑃 multiplied by one minus 𝑃 over 𝐿. And this is known as the logistic
differential equation for population growth.

In this model, 𝑘 represents the
growth rate of the population and 𝐿 represents the carrying capacity. Notice that if the population is
small compared to the carrying capacity, then 𝑃 over 𝐿 will be close to zero. One minus 𝑃 over 𝐿 will be close
to one. And hence, d𝑃 by d𝑡 will be
approximately 𝑘𝑃. And we have this simple model for
population growth.

So initially, when 𝑃 is small
relative to 𝐿, the exponential model for population growth is appropriate. However, as the population
approaches its carrying capacity, 𝑃 over 𝐿 will tend to one. Therefore, one minus 𝑃 over 𝐿
will tend to zero. And hence, the rate of change of
the population, d𝑃 by d𝑡, will also tend to zero. If the population exceeds the
carrying capacity, then 𝑃 over 𝐿 will be greater than one, which means that one
minus 𝑃 over 𝐿 will be negative. And so the rate of change of the
population will also be negative, meaning that the population is decreasing, because
it has exceeded the maximum population that the environment can support.

Now let’s consider how we can solve
this logistic equation. Although it’s more complicated than
the model for simple population growth, it is still a separable differential
equation. We can rewrite the right-hand side
as 𝑘𝑃 multiplied by 𝐿 minus 𝑃 over 𝐿 and then separate the variables, to give
𝐿 over 𝑃 multiplied by 𝐿 minus 𝑃 d𝑃 is equal to 𝑘 d𝑡. We then solve by integrating both
sides of this equation.

Now the right-hand side will be
straightforward. But to integrate the left-hand
side, we need to use partial fractions. We can express 𝐿 over 𝑃
multiplied by 𝐿 minus 𝑃 as 𝐴 over 𝑃 plus 𝐵 over 𝐿 minus 𝑃. We can then multiply both sides of
this equation by 𝑃 𝐿 minus 𝑃 to give 𝐿 equals 𝐴 multiplied by 𝐿 minus 𝑃 plus
𝐵𝑃. We can then substitute values of
our population 𝑃 to determine the values of the constants 𝐴 and 𝐵.

When 𝑃 is equal to zero, we find
that 𝐿 is equal to 𝐿𝐴. And therefore, 𝐴 is equal to
one. When the population 𝑃 is equal to
𝐿, we find that 𝐿 is equal to 𝐵𝐿. And therefore, 𝐵 is also equal to
one. So the fraction 𝐿 over 𝑃
multiplied by 𝐿 minus 𝑃 can be expressed as the partial fractions one over 𝑃 plus
one over 𝐿 minus 𝑃.

Substituting back into our
integral, we’re now able to perform this integration. We recall that the integral of one
over 𝑥 with respect to 𝑥 is equal to the natural logarithm of the absolute value
of 𝑥 plus some constant of integration 𝑐. And if we were to use the method of
substitution, we’d see that the integral of one over 𝑘 minus 𝑥 with respect to 𝑥
for some constant 𝑘 is equal to negative the natural logarithm of the absolute
value of 𝑘 minus 𝑥 plus some constant 𝑐.

So we have that the natural
logarithm of the absolute value of 𝑃 minus the natural logarithm of the absolute
value of 𝐿 minus 𝑃 is equal to 𝑘𝑡. And we’ve just included one
constant of integration 𝑐 on the right-hand side.

Now the population 𝑃 is greater
than zero. And as we need the population to be
less than the carrying capacity 𝐿 in order for this model to be meaningful, 𝐿
minus 𝑃 is also greater than zero. So the absolute value of 𝑃 is just
𝑃 and the absolute value of 𝐿 minus 𝑃 is 𝐿 minus 𝑃. We therefore have the natural
logarithm of 𝑃 minus the natural logarithm of 𝐿 minus 𝑃 equals 𝑘𝑡 plus 𝑐.

Multiplying through by negative one
and then applying laws of logarithms gives that the natural logarithm of 𝐿 minus 𝑃
over 𝑃 is equal to negative 𝑘𝑡 plus some constant 𝑐 two. We can then raise 𝑒 to the power
of each side, knowing that this will cancel out the natural logarithm on the
left-hand side, giving 𝐿 minus 𝑃 over 𝑃 equals 𝑒 to the power of negative 𝑘𝑡
plus 𝑐 two. And using laws of exponents, we can
express this as 𝐴𝑒 to the power of negative 𝑘𝑡.

We can then rewrite the fraction on
the left-hand side as 𝐿 over 𝑃 minus 𝑃 over 𝑃, which simplifies to 𝐿 over 𝑃
minus one. And then add one to each side of
the equation to give 𝐿 over 𝑃 is equal to 𝐴𝑒 to the negative 𝑘𝑡 plus one. Finally, we can multiply both sides
of the equation by 𝑃 and then divide by 𝐴𝑒 to the negative 𝑘𝑡 plus one to give
𝑃 equals 𝐿 over one plus 𝐴𝑒 to the negative 𝑘𝑡. And this is the general solution to
the logistic model for population growth.

Now we can also determine the value
of the constant 𝐴 if we’re told the population at time zero. If 𝑃 equals some value 𝑃 nought
at 𝑡 equals zero, then we have the equation 𝑃 nought equals 𝐿 over one plus 𝐴𝑒
to the power of zero. But 𝑒 to the power of zero is just
one. Rearranging this equation gives an
expression for 𝐴 in terms of the carrying capacity 𝐿 and the initial population 𝑃
nought. 𝐴 is equal to 𝐿 minus 𝑃 nought
over 𝑃 nought.

And so we see the solution to the
logistic model for population growth. We can quote this as a general
result when answering questions on this topic. Notice that this time as 𝑡
approaches infinity, 𝑒 to the power of negative 𝑘𝑡 will tend to zero. And so the population will tend to
𝐿 over one, which is just 𝐿, the carrying capacity of the population. Let’s now consider some examples of
this logistic model. In our first example, we’ll see how
to write a logistic differential equation from a physical description.

Suppose a population grows
according to a logistic model with a carrying capacity of 7500 and 𝑘 equals
0.006. Write the logistic differential
equation for this information.

We’re told which type of population
growth model to use in this question. So we can quote the standard
logistic differential equation. It’s d𝑃 by d𝑡 equals 𝑘𝑃
multiplied by one minus 𝑃 over 𝐿, where 𝑘 is the growth rate of the population
and 𝐿 is the carrying capacity. We’ve been given both of these
values in the question. So we just need to substitute them
into the logistic differential equation.

We have then that d𝑃 by d𝑡 is
equal to 0.006 — that’s 𝑘𝑃 — multiplied by one minus 𝑃 over 7500 — that’s 𝐿. We’re only asked to write the
logistic differential equation. So there’s no need for us to
attempt to solve this and we’re done.

In our next example, we’ll see how
to find the particular solution for a given logistic differential equation with an
initial condition.

Suppose a population’s growth is
governed by the logistic equation d𝑃 by d𝑡 equals 0.07𝑃 multiplied by one minus
𝑃 over 900, where 𝑃 of zero is equal to 50. Write the formula for 𝑃 of 𝑡.

Writing 𝑃 of 𝑡 means that we need
to find the solution to this logistic equation. We can begin by writing down its
general form. We know that for the logistic
equation d𝑃 by d𝑡 equals 𝑘𝑃 multiplied by one minus 𝑃 over 𝐿 that its solution
is given by 𝑃 equals 𝐿 over one plus 𝐴𝑒 to the negative 𝑘𝑡, where 𝐴 is equal
to 𝐿 minus 𝑃 nought over 𝑃 nought. Here 𝑘 represents the growth rate
of the population. 𝐿 is the carrying capacity. And 𝑃 nought is the initial
population. We can identify each of these
values from the information given in the question.

First, we see that 𝑘 is equal to
0.07 and 𝐿 is equal to 900. We’re also told that 𝑃 zero is
equal to 50. So we can fill in each of the
values in the general solution. Let’s work out 𝐴 first of all. 𝐴 is equal to 𝐿 minus 𝑃 nought
over 𝑃 nought. That’s 900 minus 50 over 50 or 850
over 50, which is equal to 17.

Now we can substitute into the
general form of the solution. 𝑃 is equal to 𝐿 — that’s 900 —
over one plus 𝐴 — that’s 17 — 𝑒 to the power of negative 𝑘 — that’s negative 0.7
— 𝑡. So we have our solution for 𝑃 or
𝑃 of 𝑡. It’s equal to 900 over one plus
17𝑒 to the power of negative 0.07𝑡.

In our final example, we’ll see how
to calculate the growth rate of the population and then calculate the population at
a given time 𝑡.

Suppose a population grows
according to a logistic model with an initial population of 1000 and a carrying
capacity of 10000. If the population grows to 2500
after one year, what will the population be after another three years?

We know that the general solution
to the logistic model is given by 𝑃 of 𝑡 is equal to 𝐿 over one plus 𝐴𝑒 to the
negative 𝑘𝑡, where 𝐴 is equal to 𝐿 minus 𝑃 nought over 𝑃 nought. Here 𝐿 is the carrying capacity of
the population, 𝑃 nought is the initial population, and 𝑘 is the growth rate of
the population.

We’ve been given some of this
information in the question. We’re told that the initial
population, 𝑃 nought, is 1000. And we’re told that the carrying
capacity, 𝐿, is 10000. But we haven’t been given the
growth rate of the population. Instead, we’ve been given another
pair of values for 𝑃 and 𝑡. We’re told that the population
after one year is 2500. We’ll be able to combine this
information with our values of 𝐿 and 𝑃 nought in order to determine the growth
rate of the population.

First, we can work out the value of
the constant 𝐴. It’s 𝐿 minus 𝑃 nought over 𝑃
nought, 10000 minus 1000 over 1000, which is 9000 over 1000, which is nine. Substituting 𝐿 and 𝐴 into our
model then, we have that 𝑃 of 𝑡 is equal to 10000 over one plus nine 𝑒 to the
power of negative 𝑘𝑡.

Now we can use the population after
one year in order to determine the value of 𝑘. Substituting 2500 for 𝑃 and one
for 𝑡, we have 2500 equals 10000 over one plus nine 𝑒 to the negative 𝑘. To solve for 𝑘, we first multiply
by one plus nine 𝑒 to the negative 𝑘 and then divide by 2500, giving one plus nine
𝑒 to the negative 𝑘 is equal to four. We can then subtract one and divide
by nine, giving 𝑒 to the negative 𝑘 equals four minus one over nine, which is
three-ninths or one-third.

We then take natural logarithms of
each side, knowing that this will cancel out with the exponential on the left-hand
side, to give negative 𝑘 equals the natural logarithm of one-third. We can then multiply by negative
one to give 𝑘 equals negative the natural logarithm of one-third. And using laws of logarithms, this
is equal to the natural logarithm of three. So we found the value of 𝑘, the
growth rate of the population. Our model therefore becomes 𝑃 of
𝑡 equals 10000 over one plus nine 𝑒 to the negative 𝑡 multiplied by the natural
logarithm of three.

Now we’re asked for the population
after another three years, which means we’re looking for the population when 𝑡 is
equal to four. So the final step is to substitute
𝑡 equals four into our model. We have then that 𝑃 of four is
equal to 10000 over one plus nine 𝑒 to the negative four ln three. This actually works out very
nicely, which you can see if you apply laws of logarithms in the denominator. We get 10000 over 10 over nine,
which is 10000 times nine over 10, which is equal to 9000. So we find that the population
after another three years, that is, the population four years after we’ve started,
is 9000.

Let’s now summarise what we’ve seen
in this video. We’ve introduced the logistic
model, which can be used as a more realistic model of population growth to reflect
the fact that populations usually cannot grow without limitation due to finite
resources. The logistic differential equation
for population growth is d𝑃 by d𝑡 equals 𝑘𝑃 multiplied by one minus 𝑃 over 𝐿,
where 𝑘 is the growth rate and 𝐿 is the carrying capacity.

By separating the variables and
integrating, we’ve found that the solution to the logistic differential equation
with an initial population of 𝑃 nought is 𝑃 of 𝑡 equals 𝐿 over one plus 𝐴𝑒 to
the negative 𝑘𝑡, where 𝐴 is equal to 𝐿 minus 𝑃 nought over 𝑃 nought. This produces a much more realistic
model of population growth, where the population initially grows at an exponential
rate. But then the rate of the growth
slows down. And finally, the population levels
off at the carrying capacity of the population.