- [Lecturer] Welcome to a second lesson on mixture problems
involving first order differential equations.
In this video, we'll focus on
different inflow and outflow rates.
So if we let A of T be the amount of substance
dissolved in the liquid of a tank at any time T,
where liquid is entering and leaving the tank,
and the liquid entering the tank will have a different
concentration of the substance dissolved in it.
So if this is the case, we can model
the change of the mixture using the differential equation
DA/DT equals R sub one minus R sub two
where DA/DT is a rate at which A of T is changing
in the tank, R sub one is a rate at which
A of T is entering the tank and R sub two
is a rate at which A of T is leaving the tank.
To get a visual of this...
We have a certain concentration being pumped
into our mixing tank and then pumped out.
And for our example, the inflow rate and outflow rate
will be different and because they're different,
notice how the inflow rate is going to be higher
than the outflow rate, the amount of liquid
in the tank is going to be increasing.
So we will have to be considering this
when setting this up.
For our example, a large tank holds 300 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 three gallons per minute.
How much salt is in the tank after 12 minutes?
So we're going to start with our differential equation DA/DT
equals R sub one minus R sub two.
And we'll start by finding R sub one and R sub two,
where again, R sub one is the rate at which A of T
is entering the tank and R sub two the rate at which
A of T is leaving the tank.
Well R sub one is pretty straightforward.
We have two pounds per gallon pumped at a rate of
four gallons per minute into the tank.
So R sub one is going to be product of these two.
Now I'm writing the units in fraction form
so we can see that the units of the gallons simplifies out.
So R sub one is going to be eight pounds per minute.
Now R sub two is going to be a little bit trickier
because for the concentration,
the amount of salt is changing
and so is the amount of gallons in the tank.
So for the concentration, we're going to take
the amount of salt and divide by
the number of gallons in the tank.
So A or A of T is the amount of salt in the tank.
And the number of gallons is not just 300 because
notice how the solution is being pumped in at
four gallons per minute
and pumped out at three gallons per minute.
So after every minute, we have one more gallon in the tank.
So the total gallons is going to be 300 plus 1T,
where T is the number of minutes.
Again, because we have one extra
gallon every minute, we have 1T here.
This is going to be pounds per gallon
times the rate of three gallons per minute.
Again, gallons simplifies out.
So R sub two is going to be 3A
divided by 300 plus T and this would be pounds per minute.
So the differential equation, to model this situation
that we need to solve is DA/DT equals
R sub one, which is eight minus R sub two,
which I'm going to go ahead and write as three
divided by the quantity 300 plus T times A.
The reason I'm writing it this way is
it's easier to recognize that this is a linear
first order differential equation.
And also, we know that A of zero is equal to 40
because that at time T equals to zero,
there's 40 pounds of salt in the tank.
Now we'll go ahead and solve this
using an integrated factor.
So the first step would be to write this in standard form.
Then, for a quick review, you may want to pause the video
on this slide to review how to solve a linear
using the integrating factor given here
where P of X is equal to the coefficient of our
function Y, or in our case, the function A.
So write this in standard form.
We're going to go ahead and add this quantity here
to both sides of the equation.
So we'd have DA/DT
plus three divided by 300 plus T
times A equals eight.
So here's our function P of T
that we'll use to find the integrating factor.
So the integrating factor, we'll call it mu of T,
is going to be equal to E raised to the power of the
integral of this function here.
Three divided by 300 plus T, DT.
So to integrate this, we'll go ahead and let U equal
the denominator of 300 plus T and notice that
DU is equal to DT.
So this just ends up being E raised to the power of,
this would be three times natural log
at the quantity 300 plus T.
We know T's always positive.
So we don't need absolute values here.
And from here, we can use the property
of logarithms to move this three
here to the position of our exponent.
So now we have E raised to the power of natural log
times the quantity 300 plus T to the third.
And this simplifies very nicely because we have base E here
and this is natural log or log base E.
So when we're integrating factors,
it's just the quantity 300 plus T raised to the third.
Now we'll take this form of our differential equation
and multiply everything by integrating factor.
Now remember when we do this, the left side of
this differential equation or this side here
is equal to the derivative of the product
of the integrating factor in our function A.
So we can say the left side is equal to the derivative
of our integrating factor, which is 300 plus T to the third
times our function A or A of T.
And we can verify this using the products rule,
but if the derivative is equal to the left side
of this equation, it must also equal to the right side.
So this is equal to eight times the quantity
300 plus T raised to the third.
And from here, we can now integrate both sides
of this equation with respects to T
and then solve for our function A.
So now we're going to integrate the left side
and integrate the right side.
Well, on the left side,
the integral derivative undo each other.
So we're left with the quantity 300 plus T to the third
times A and then on the right side, if we integrate this,
we would have eight times 300 plus T to the fourth
divided by four plus C.
Notice how the eight and the four simplify nicely.
Let's continue this on the next slide.
Now to solve for A, we'll divide by this quantity here.
So on the left side, this simplifies to A, or A of T.
On the right side, notice how all of this simplifies
to just one factor of 300 plus T.
We'll go and distribute the two.
That would give us 600 plus 2T plus
here we'll go ahead and move this up into the numerator,
which would change the sign of the exponent.
So we'd have C times 300 plus T to the negative third.
This would be the general solution
to our differential equation.
But remember, we also know that A of zero is equal to 40.
So this would allow us to find our constant here
and find our particular solution
for this differential equation.
So if A of zero is equal to 40,
we'll substitute zero for T and then set
this equal to 40.
So we would have 600.
2T would be zero.
So we'd have 600 plus C times,
this would just be 300 to the negative third power
and this equals 40.
So next, we would subtract 600 on both sides.
That would give us negative 560.
And here, we could divide by 300 to the negative third
or just multiply it by 300 to the positive third.
And from here, we'd add the exponents.
That would give us 300 to the zero,
which is equal to one.
So we'd have C equals this product here,
which we'll go to the calculator for.
We have negative 1.512 times 10 to the 10th.
Using decimal notation, we'd have negative 1512.
We need seven more zeroes.
So we have negative 15 billion 120 million
for our constant C, which we'll now substitute back
into our function A of T for C.
So we have A of T equals 600 plus
2T minus 15 billion 120 million
times 300 plus T to the negative third power.
So now to answer the question, how much salt is in the tank
after 12 minutes, we need to find the value of A of 12.
So we'll substitute 12 for T.
This would be 600 plus 24, 624,
minus 15 billion 120 million times,
this would be 312 to the power of negative three.
And we'll go back to the calculator.
So after 12 minutes, there's approximately
126.16 pounds of salt in the tank.
Now it doesn't ask, but remember, because the inflow
and outflow rates are different by one gallon per minute,
after 12 minutes, it would be 312 gallons in the tank.
Okay, I hope you found this helpful.