- [Narrator] Welcome to a lesson on mixtures
involving first order differential equations.
In this video we'll look at an example
where the inflow and outflow rates are the same.
So if we let A of t be the amount of the substance dissolved
in a liquid in a tank at any time t.
Where liquid is entering and leaving the tank.
The liquid entering the tank
will contain a different concentration
of the substance dissolved in the tank.
So what we're saying is we have a tank
that contains a certain concentration of a solution.
A different concentration is entering the tank.
Which instantly mixes with the current solution
and then the new solution is pumped out of the tank.
So if this is the case we can say that
dA dt is equal to, R sub 1, minus R sub two.
Where dA dt is the rate at which
A of t is changing in the tank.
R sub 1 is the rate at which A of t is entering the tank.
And R sub 2 is the rate at which A of t is leaving the tank.
So let's take a look at our example.
A large tank holds 200 gallons of brine solution
with 40 pounds of salt.
A concentration of two pounds per gallon is pumped in
at a rate of four gallons per minute.
The concentration leaving the tank
is pumped out at a rate of four gallons per minute.
The concentration leaving the tank
is also pumped out at a rate of four gallons per minute.
How much salt is in the tank after one hour?
And how much salt is in the tank after a very long time?
So using our differential equation
dA dt equals, R sub 1, minus R sub 2.
So we want to start this by finding R sub 1.
The rate at which A of t is entering the tank
and R sub 2 the rate at which A of t is leaving the tank.
To find these we'll have to multiply
the concentration times the rate.
So for R sub 1 notice how there are two pounds per gallon,
but it's being pumped in at four gallons per minute.
So we need to find this product.
So we have two pounds per gallon,
times four gallons per minute.
Now the reason I'm writing this in fraction form
is so that we can easily see
that the units of gallons simplifies out.
Therefore R sub one is going to be eight pounds per minute.
Now R sub two is going to be a little bit trickier.
We're going to have to write an expression
for the concentration,
because it's changing as the mixing occurs.
Notice how we start off with 40 pounds
of salt per 200 gallons,
but the 40 is changing
so we can't use that for a concentration.
We have to use an expression.
So we'll use A, where A is the unknown, divided by 200.
And this would be pounds per gallon.
And then the rate is the same as four gallons per minute.
Again notice how gallons simplifies out.
So we're going to have, this would be 4 A over 200.
And this would be pounds per minute.
But this does simplify four over 200 as one fiftieth.
So we have A divided by 50 pounds per minute.
So our differential equation would be
dA dt
equals eight, minus
A, divided by 50.
Well this is actually
a linear first order differential equation.
Which we can solve by using an integrating factor.
Let's go ahead and set this up on the next slide.
So just to review if we have
a linear first order differential equation
then we're going to write the differential equation
in standard form or this form here.
And then we're going to multiply everything
by our integrating factor which is given here.
So you may want to pause the video here
if you need this review.
So I'm going to go ahead and rewrite this in standard form.
So we'd have dA dt, plus one fiftieth A,
equals eight.
So in this form we should recognize
that the integrating factor u of t
is going to be equal to e raised to the power of
the integral of one fiftieth dt.
So this is just going to be
e to the t divided by fiftieth power.
Now we're going to multiply
everything in this differential equation here
by this integrating factor.
So we're going to have e to the t divided by 50 da dt,
plus one fiftieth, e, to the t divided by 50, times A,
equals eight, times e, to the t divided by fiftieth power.
Now once we multiply everything by the integrating factor.
Remember the left side of this equation or this side here
is equal to the derivative of the product
of the integrating factor and our function A.
So the derivative with respects to t,
through the integrating factor which is e,
to the t divided by 50, times our function A.
So if the left side is equal to the derivative,
then it must also equal the right side.
So we'll set this equal to eight,
e, to the t divided by the fiftieth power.
And now we'll integrate both sides of the equation
with respects to t.
On the left side the derivative
and integral undo each other.
So we're left with the exponential,
times A, must equal.
Over here we have to integrate using u substitution.
So notice here if we let u equal, t, divided by 50,
then du is equal to one fiftieth dt.
So that means 50 du is equal to dt.
So when performing the substitution we're going to have
an extra factor of 50.
So the anti-derivative will be eight, times 50,times e,
to the u, or e, to the t, divided by 50 power.
And of course plus our constant of integration.
Okay now our goal here is,
okay now our goal is to solve this for A or A of t.
So now we'll divide both sides by our exponential.
So on the left side simplifies to A or A of t.
On the right side these exponentials simplify to one.
So we have 400.
And here we can just write this as plus c,
times e, to the negative t divided by fiftieth power.
Now that we have the general solution
we can find the particular solution
using the initial condition that A of zero
is equal to 40, because there's 40 pounds of salt
to begin with in the 200 gallons.
Let's go ahead and do this on the next slide.
Okay so if A of zero equals 40,
we'll replace t with the zero,
we would have 400 plus c, times e, to the zero,
which is equal to one.
So this is just c.
This must equal 40.
So we'll subtract 400 on both sides.
So c must equal, this will be negative 360.
Therefore the function that models the mixing solution
is A of t,
equals 400, minus 360, times e,
to the power of negative t divided by 50.
And now we can determine how much salt
is in the tank after one hour.
But we need to be careful here because t is time in minutes.
Therefore for one hour we have to find A of 60.
Not A of one.
And we know this because the initial flow rates
were given in gallons per minute.
So our exponent here would be negative 60, divided by 50.
And now we'll go to the calculator.
So it's approximately 291.6 pounds of salt in the tank
after one hour.
And now for the last question.
How much salt is in the tank after a very long time?
We can answer this by finding the limit
as t approaches infinity of A of t.
Which would be again 400 minus 360, times e,
to the negative t, over 50 power.
But I'm going to go ahead and make this a fraction.
Move this down to the denominator to help us find the limit.
And we should recognize it as t approaches infinity.
Where it gets larger and larger.
So does this denominator and therefore this approaches zero.
Therefore, the limit is equal to 400.
This would be the amount of salt in the tank
after a very long time.
And hopefully this makes sense,
because remember we have 200 gallons in the tank.
And since the inflow and outflow rates are the same
eventually the concentration would eventually be the same
as the inflow concentration which was two pounds per gallon.
Again gallons simplifies out.
Leaving us with 400 pounds of salt after a very long time.
Okay I hope you found this helpful.