Video Transcript
An aquarium contains fish with a
carrying capacity of 1200 and a growth rate of 8 percent. If the initial population of the
fish is 400, what is the population of the fish at any given time?
In this question, we need to
consider models for population growth. But one of the key features is that
this aquarium has a carrying capacity of 1200, which means that this population of
fish cannot grow in an unrestricted way. There is a maximum population of
fish that can be supported in this aquarium. Now, this tells us that instead of
using the simple model for population growth, we instead need to use the logistic
model for population growth, which incorporates this feature that there is a maximum
population that can be supported.
This logistic model, which is the
solution to the logistic differential equation for population growth, has the
following general form: the population at time 𝑡 is given by 𝐿 over one plus 𝐴𝑒
to the power of negative 𝐾𝑡. Where 𝐿 represents the carrying
capacity of the population, 𝐾 is its growth rate, and 𝐴 is equal to 𝐿 minus 𝑃
nought over 𝑃 nought, where 𝑃 nought is the initial population. We can determine several of these
values from the information given in the question. We’re told that the carrying
capacity — so that’s 𝐿 — is 1200. We’re also told that the growth
rate is 8 percent. So, as a decimal, that’s 0.08. And we’re told that the initial
population of the fish is 400. So, that’s our value for 𝑃
nought.
We can, therefore, calculate the
value of the constant 𝐴. It’s 𝐿 minus 𝑃 nought over 𝑃
nought, so 1200 minus 400 over 400. That’s 800 over 400 which is equal
to two. To find the population of fish at
any given time then — so that’s an expression for the population 𝑃 in terms of 𝑡 —
we just need to substitute the values of 𝐿, 𝐴, and 𝐾 into the logistic model. Doing so gives that the population
at time 𝑡 is equal to 1200 over one plus two 𝑒 to the power of negative
0.08𝑡. Now, this would be a perfectly
acceptable form in which we leave our answer. But we may prefer to deal with
positive exponents for our exponential rather than negative.
We can achieve this by multiplying
both the numerator and denominator of our fraction by 𝑒 to the power of positive
0.08𝑡. And as we’ve multiplied both the
numerator and denominator by the same, we’ve created an equivalent expression. Doing so gives 1200𝑒 to the power
of 0.08𝑡 in the numerator. And in the denominator, we have one
multiplied by 𝑒 to 0.08𝑡. And then, we have two 𝑒 to the
negative 0.08𝑡 multiplied by 𝑒 to the power of 0.08𝑡. So, we recall here that we add the
exponents together, giving two 𝑒 to the power of zero. But of course, 𝑒 to the power of
zero is just one. So, this simplifies to two.
So, by using the logistic model for
population growth, we found that the population of fish at any given time in this
aquarium, which has a carrying capacity of 1200, a growth rate of 8 percent, and an
initial population of 400, is given by 𝑃 of 𝑡 equals 1200𝑒 to the power 0.08𝑡
over two plus 𝑒 to the power of 0.08𝑡. We can find the population any
given time by substituting the relevant value of 𝑡.
