- [Narrator] Welcome to the first of several videos
on the applications of first order differential equations.
This video will focus on exponential growth.
The differential equation dP/dt equals k times P
or P prime of t equals k times P of t
models exponential growth,
which means the rate of growth
is proportional to the population or amount.
So looking at these again, this is telling us
the change of P with respects to t
is equal to some constant k times P,
or the derivative is equal to a constant
times the original function P of t.
And this can also be used to model continuous interest.
While we will solve this differential equation, find P of t,
the only function that satisfies this differential equation
is P of t equals P sub zero times e
raised to the power of k t.
We should recognize this exponential function
from our algebra class
where P sub zero is the initial amount,
k is the growth rate and t is the time.
But let's start by looking at an example.
A population of a small town grows
proportional to its current population.
The initial population is 5000 and grows 4% per year.
This can be modeled by the differential equation
dP/dt equals 0.04 times P,
and P of 0 is equal to 5000.
We want to find an equation to model the population,
determine the population after three years,
then determine how long it will take
for the population to double.
So to find the equation to model the population,
we're going to solve this differential equation.
So if we write this in differential form,
we would have dP
is equal to 0.04 P dt.
We want the P on the left side of the equation
so we're going to multiply both sides by one over P.
So we'd have one over P dP
is equal to 0.04 dt.
Now we can integrate both sides of this equation.
The antiderivative of one over P
would be the natural log absolute value of P.
But we know P has to be positive
because it's the population.
So we have the natural log of P.
We'll put the constant of integration on the right side.
So this is going to be equal to
the antiderivative of 0.04 with respects to t
would be 0.04 t
plus the constant of integration.
Now once we have this equation, if this is true,
then e raised to the power of natural log P
must also equal e raised to the power of this sum.
By doing this, the left side simplifies very nicely to P,
which we know is going to be a function of t.
So we'll write P of t equals...
Now because we have a sum here,
this would be e to the 0.04 t
times e to the C.
But e to the C is just a constant.
So let's let e to the power of C equal P sub zero.
So now we have an exponential function
that would satisfy this differential equation.
It's going to be P of t equals P sub zero
times e raised to the power of 0.04 t,
which does fit the form of the exponential function
that we discussed on the previous slide, also given here.
But now let's go ahead and verify
that P sub zero would actually be 5000.
So P of zero would be equal to
times e to the zero power,
which again we know must equal 5000.
Well e to the zero is equal to one,
so we do have P sub zero equals 5000.
Giving us our particular solution
to the differential equation,
we have P of t is equal to 5000
times e raised to the power of 0.04 t.
Again which matches the exponential function
that we knew would be our solution given here.
Now let's go ahead and use this function
to answer a couple questions.
We want the population after three years.
So we're just going to substitute three for t.
You need a calculator for this part.
So 5000 times e
raised to the power of 0.04 times three.
So the population will be approximately,
we'll round to the nearest person,
so we're going to round down: 5,637 people.
The last question asks us to determine
how long it will take for the population to double.
So again, starting with our population function,
the initial population is 5000.
5000 times two is 10,000.
So we're going to set P of t equal to 10,000 and solve for t.
Divide both sides by 5000.
This would be one, 10,000 divided by 5000 is two.
This will always be two
when trying to find the doubling time.
And to solve for t, we'll have to use
the properties of logarithms.
We'll take the natural log of both sides of the equation.
So on the left side we have natural log two equals-
Here we can apply the power property of logarithms.
We can take this exponent here and move it to the front.
So we have 0.04 t times natural log e.
But natural log e is equal to one, so that simplifies out.
So now we just divide by 0.04
and go to our calculator to find
the approximate doubling time.
So we have natural log two divided by 0.04,
which happens to be our growth rate.
So the doubling time is approximately, let's say 17.3 years.
I do want to point out there's a shortcut
for determining doubling time.
Let's call it capital T.
The doubling time is always going to be equal
to natural log two divided by the growth rate, we'll call K.
Okay, I think we're short on time, so I'll go ahead
and take a look at the second example in part two.
I hope you found this helpful.