1520 9 5 4 Mixing Problem HD 1080 WEB H264 4000

Let's take a look at a mixing problem.

In this example, we have a tank

that initially (at our starting time)
holds 100 liters of pure water.

And what we've got coming in is,
we've got brine that has salt in it,

and the concentration of salt is
0.5 kilograms of salt for every liter,

and that's coming into the tank
at a rate of 4 liters per minute.

The big assumption
we're making here —

and this, in real life,
is not 100% realistic —

but we're assuming that as soon
as that brine enters the tank,

it is immediately mixed perfectly
in with whatever is already in there.

Then that mixture is draining out
at a rate of 3 liters per minute.

Now, what makes this different

from the mixing problem that we saw
in the separable equation section is,

the rate of liquid leaving the tank
isn't the same as the rate coming in.

So in one of the worksheet
problems from a previous section,

we had the liquid leaving
the tank at the same rate,

and when we did that,

we ended up with a
separable differential equation.

In this case, because the liquid is
leaving the tank at a different rate,

we're going to end up with a linear
differential equation instead.

So one thing we need is,

we have to figure out
what we're even solving for.

And basically what we want
to do is, we want to find y of t.

And what that's going to represent is,

it's going to be the amount
of salt in the tank at time t.

After t minutes have gone by,
how much salt is in the tank?

One thing we know right away

is that the tank initially
had pure water in it.

That tells us our initial value is 0.

There's no salt in the tank
at time 0; it's pure water.

One other thing that's
going to happen with this is,

every minute that goes by,

think about the amount
of liquid in this tank.

If it was leaving at the same rate
that it was coming in,

that would stay constant.

But every minute that goes by,

one extra liter is coming in
that doesn't get drained out.

The total amount of liquid in the tank
is actually changing as time goes by.

So it's not going to stay 100 liters.

The amount of liquid in the tank

is going to be 100 liters plus
1 liter for every minute that goes by.

Since t represents
the number of minutes,

I'm just going to write that as 100 plus t,

and that's the number of liters of liquid

that are going to be in
this tank at any given time.

Now, in real life, eventually,

we'd reach a point where
this process can't continue

because the tank only has physically
enough room for a certain amount.

That's something to bear in mind.

But for our purposes,

we'll assume that we're
early enough in the process

that this model will
make sense and will hold.

So one thing that we found out
in previous sections is,

if we want to know what dy [over] dt is

(that would be the change
in the amount of salt),

we have to figure out what rate
salt is coming into the tank,

and then we have to subtract
the rate that salt is leaving the tank.

And the rate that salt is coming in

is actually pretty
straightforward to calculate.

Each liter that comes
in has half a kilogram (0.5 kg)

and there's 4 liters
coming in every minute.

If we just multiply those
two numbers together,

we'll end up with 0.5
times 4, which is 2.

If you actually look at units here,
it's kilograms per liter,

and then we're multiplying
that by 4 liters per minute.

The liters actually cancel

and we're going to end up
with 2 kilograms perminute.

So that's the rate of salt
coming into the tank.

Every minute,
2 kilograms of salt come in.

Now, initially, basically
no salt is going to leave

because the water at the beginning
is almost purely water.

But as it gradually gets saltier,

then some of that salt
is going to leave the tank

when this brine is drained.

In order to figure out the rate out

(this is the one that takes
a little bit more thinking),

we have to figure out

the concentration of
salt in the current liquid,

and then we have to multiply
that by the rate that it's leaving,

which is 3 liters per minute.

Well, the concentration of salt

is just the number of kilograms of salt
divided by the total amount of liquid.

We know that there are y kilograms
in the tank at any given time

and we could divide that
by the amount of liquid,

which is 100 plus t liters,

and that would tell us
our concentration:

it's y kilograms per 100 plus t liters.

This is the biggest difference

between what we're
doing in this section

compared to when
these two rates match.

If these were the same rate,
we could just say y divided by 100

because the amount of liquid
in the tank would stay constant.

But because of this extra
t that gets added in,

this is what's going to
make it no longer be a—

It's no longer going to
be a separable equation.

It'll be a linear equation.

So we've got our concentration,
which is y over 100 plus t

(technically, that is kilograms per liter),

and it's exiting at a rate
of 3 liters per minute,

so liters are going to cancel.

And once again, this is going
to be kilograms per minute.

That's the same units
that the rate in was.

So we can subtract, multiply these,
we'll get 3y over 100 plus t.

This is the differential equation
that we're going to try to solve.

This actually is a linear differential
equation, and to see that,

we actually might want to
rearrange the terms a little bit.

In the standard form for
a linear differential equation,

dy [over] dt is by itself.

Then if I move this over,

instead of writing it
as 3 over 100 plus t,

I'm going to write it as
3 over 100 plus t, multiplied by y,

and then that's going to equal 2.

So in this form, dt is by itself.

This function right here
is playing the role of P.

It's not a function of x,
but it's a function of t.

That's our independent variable this time,
and then that's playing the role of Q.

Just like what we've done
for the other examples,

we're going to use this to
find an integrating factor.

We're going to multiply everything
by that integrating factor,

and that should turn the left side
of the equation into a product rule.

That's something that we can invert
and then eventually solve for y.

All right, so here's our first order
linear differential equation.

It is already in standard form
because we did that already,

and that function right there
is playing the role of P.

This time, it's a function
of t instead of x,

it’s 3 over 100 plus t,

and our integrating factor

is always given by e to the
integral of the function P,

which in this case is e to the
integral of 3 over 100 plus t.

And like the previous example,
we can take that 3 out if we want.

If you want to go through
the full steps here,

this is-- technically could be
done by using a substitution.

But if u is 100 plus t,
du is just 1 times dt.

We don't have to worry about
an extra factor appearing here,

and this integral is just going to end up
being the natural log of 100 plus t.

In this example, we're assuming that
t is positive since it represents time.

Also, 100 plus t was
the total number of liters

in the tank at any given time,

so we are assuming that
that will never be negative.

We won't have a negative
amount of liquid in the tank.

Like we did before,

we can utilize that property of logarithms

that lets us take this coefficient

and turn it into a power
within the logarithm itself.

It will be the natural log
of 100 plus t cubed.

And e and natural log are inverses,
so once those cancel out,

that tells us that the integrating
factor we need to use

is 100 plus t to the 3rd power.

Now the next step is going to be

to multiply everything
by that integrating factor.

And if we did it right, the left side of
the equation should be a product rule.

We've got our integrating
factor times dy [over] dt.

and when we multiply our next thing
by the integrating factor,

we're going to have
3 times 100 plus t cubed,

but that gets divided by 100 plus t.

So that's just going to become
3 times 100 plus t squared

because one of the
100 plus t’s from the top

will cancel this one on the bottom.

Then the right side of the equation

becomes 2 times 100
plus t to the 3rd.

The whole purpose
of the integrating factor

is to try to make the left side of
the equation become a product rule.

If the integrating factor itself is f,
then that really is f prime.

And if y is g, dy [over] dt is g prime.

That means this
really is a product rule.

This is the derivative
with respect to t of f times g

and f is 100 plus t to the 3rd

and g is just y, and that's equal
to the right side of the equation.

Now that we've got our product rule
on the left side of the equation,

we can integrate it to make
the derivative go away,

so integrate with respect to t,

and that will equal the integral
of the right side of the equation,

also with respect to t.

The derivative of an integral cancels out.

That leaves us with 100
plus t to the 3rd times y.

And then for this integral,

this is another one where you
could do a full U-substitution.

If u is 100 plus t, then du is just dt.

The antiderivative of
something to the 3rd,

the power just goes up by one
and then you divide by 3.

Since there's no chain rule
issue with the inside

(the derivative of the inside is 1),

you basically just treat this
as if it was our variable.

We're going to end up with 2

and then that's going to be

times one fourth [1/4]
100 plus t to the 4th power,

and then we're going to have our constant.

But instead of running 2 times 1/4,
let's just call that one half [1/2].

Now, if we wanted a general solution,

we would divide everything
right now by 100 plus t to the 3rd.

But because we have
an initial condition here,

we can actually find
our constant right now

or you can get the general solution
first and then solve for C.

It does not matter, but in this case,

I've already got it almost
to the form C equals,

so let's just use this version.

We were told the initial amount
of salt in the tank was 0.

So when t is 0, y is also 0.

We're going to have 100 plus t.
[corrects self] Oops, t is 0,

times y, which is also 0,

and that's going to equal
1/2 of 100 plus 0 to the 4th.

That's what C is.

This is all 0, that goes away.

And 100 to the 4th is—

basically, we just need to keep
adding zeros to this thing,

so let's think about that.

100 to the 4th would be 100

and then two more zeros is 100 squared.

Two more zeros is 100 to the 3rd,

and then two more zeros
is 100 to the 4th.

That's a lot of zeros,
and then that gets multiplied by 1/2.

It's 1/2 of, let's see, 100 million.

so that's what C is.

That means that C is half of that,
so it's 50 million.

You can even leave it as
1/2 of 100 to the 4th if you prefer,

but that's what this ends up being.

So that's our constant.

And now that we know what that is
(so we just figured out C),

now we can solve the equation for y.

For space reasons,
I'm going to try to fit it in here.

We're going to have y equals,

and if we take this expression right here

and we divide by
100 plus t to the 3rd,

then all but one of the
100 plus t’s will cancel.

We'll have 1/2, 100 plus t.

Then the constant
we got was 50 million

and that is going to be divided
by this 100 plus t to the 3rd,

or we could write it as multiplying it
by 100 plus t to the negative 3rd,

and that would be okay as well.

But what this equation tells us is,

it tells us how much salt in kilograms
is going to be in our tank at any time,

at least as long as
this model is realistic.

As the tank fills up,

eventually will reach a point
where this won't work anymore,

but it should describe what happens

for at least the first
however-many minutes

that we want to know
something about it.

That would be the solution
to this differential equation.