Welcome to an example of a mixture problem
that can be solved using a separable differential equation.
A tank contains 1,080 liters of pure water.
Solution that contains 0.05 kilograms of sugar per liter
enters the tank at a rate of three liters per minute
and is thoroughly mixed.
The new solution drains out of the tank at the same rate.
Part a, we're asked how much sugar is
in the tank at the beginning?
And we're told the tank begins with pure water
and therefore there are zero kilograms of sugar
in the water at time t equals zero.
Notice how this also tells us that the function
y or y of t is equal to the amount
of sugar in the tank at time t.
Part b we're asked to find the amount of sugar
after t minutes, given by the function y of t.
And then for part c, as t becomes large,
what is the value of y of t approaching?
In other words, we want to find this limit here.
So we can model the situation using the differential
equation dA/dt equals R sub one minus R sub two.
In our case though, the function is going to be
y of t not a of t.
So we'll actually use a differential equation, dy/dt
equals R sub one minus R sub two.
Where dy/dt equals the rate of change
of the sugar in the tank.
R sub one equals a rate of change
of the sugar entering the tank.
And R sub two equals the rate of change
of the sugar leaving the tank.
So to find these rates of change,
we'll find the product of the concentration of sugar
in the solution and the flow rate.
So let's begin by finding R sub one,
which is the rate at which the sugar is entering the tank.
The solution entering the tank contains .05 kilograms
of sugar per liter.
So we'll have 0.05 kilograms per liter
times the flow rate, which is given as
three liters per minute.
We have the concentration of sugar and the solution
times the flow rate for the rate at which the sugar
is entering the tank.
Notice how liters simplifies out.
And therefore R sub one is equal to 0.15
again this would be kilograms per minute.
We now need to find R sub two, which is the rate
at which the sugar is leaving the tank.
So looking we'll have a concentration
of a solution times a flow rate,
with a flow rate leaving is the same as a flow rate in,
so the flow rate is three liters per minute
and the concentration of the solution leaving the tank
is going to be equal to the amount of sugar in the tank
at any time t, which is given by the function y of t,
or I should say, y, divided by the amount in the tank,
which is always 1,080.
So the amount in the tank is always constant
because the flow rate in and the flow rate out are the same.
If they were different, the amount in the tank
would actually be a function of t, not a constant.
And the units here again are kilograms per liter,
again liters simplify out,
so we're left with three y divided by
1,080 kilograms per minute, but this fraction does simplify
to y over 360 kilograms per minute.
The differential equation that models this situation
would be dy/dt equals R sub one minus R sub two
which would be 0.15 minus y divided by 360.
And now we need to solve
the separable differential equation.
Let's do this on the next slide.
We want this to be in the form of the y terms times dy
equals the t terms times dt.
So if we multiplied both sides by dt,
we'd have differential y equals
the quantity 0.15 minus y divided by 360 times dt.
So to get these two terms on the left side,
we'll divide both sides by this quantity,
so this would give us one over,
let's write this as 0.15, let's write this as
minus 1/360 y, dy equals dt,
and now we'll integrate both sides.
So we have the integral of one over this quantity,
equals the integral of dt, if we want one dt.
And now here we'll have to perform u substitution,
where u would be equal to 0.15 minus 1/360 y,
so differential u is equal to negative 1/360 dy,
so if we solve this for dy,
notice how negative 360 du equals dy.
So if it's helpful, we can think of all of this
as negative 360 one over u du,
the anti-derivative in terms of u would be
negative 360 natural log absolute value of u,
which means in terms of y,
we'd have negative 360 natural log
absolute value of 0.15 minus 1/360 y,
we would have a constant integration
but we'll go ahead and put the constant at the right,
so we'd have equals the integral of one
with respect to t is t, so we have t plus,
let's call it c sub one.
Now we need to work on solving this for y.
So for the first step let's multiply both sides
by negative 1/360, which would give us natural log
of the absolute value of 0.15 minus 1/360 y
equals negative 1/360 t and then minus 1/360 t
times c sub one, which would be just another constant,
let's go and let that constant be equal to c sub two.
So let's say, plus c sub 2,
and now I'll put the log equation
as an exponential equation,
remember natural log is log base e,
so we'd have e raised to the quantity on the right
equals this quantity here.
Or we can say 0.15 minus 1/360 y
equals e raised to the power of negative 1/360 t
plus c sub two, but now on the right side,
because we're adding exponents,
we can write this as a product,
we can write the right side as
e raised to the power of negative 1/360 t
times either the power of c sub two,
but e to the power of c sub two is just some other constant,
so let's e to the power of c sub two
be equal to c sub three.
So we can write this as 0.15 minus 1/360 y
equals c sub three times e raised to negative 1/360 t
and now again we still want to solve this for y,
so for the next step we'll subtract 0.15 on both sides,
and now we'll multiply both sides by negative 360
to solve for y.
So we have y equals negative 360 times the right side.
Let's go ahead and distribute here,
so we have y or y of t equals,
and notice for this first product we'll have
negative 360 times e sub three times e raised to this power,
but negative 360 times e sub three,
is just another constant, so let's let c be equal
to negative 360 times e sub three.
So we'd have c times e raised to negative 1/360 t
and then we'd have negative 360 times negative 0.15,
which equals positive 54, so we have plus 54.
So now we're almost there, but we need to find
the particular solution using the initial condition
that y of zero equals zero.
Remember we know y of zero equals zero
because the first question was
how much sugar's in the tank at the beginning.
Which again gave us y of zero equals zero.
So if y of zero equals zero,
we can determine the value of c
by substituting zero for t and zero for y of t.
So we'd have the equation, zero equals c times e
raised to the power of negative 1/360 times zero plus 54,
so we'd have zero equals, this is just c times e
to the zero, which is c plus 54.
So c equals negative 54.
So the particular solution to the differential equation,
which gives us the y of t for the situation is
y of t equals, again c is negative 54,
times e raised to the power
of negative 1/360 times t plus 54.
So going back to our first slide,
this is the answer for part b.
And then for part c, we want to find the limit
as t approaches infinity of y of t.
So to find this limit, it might be helpful
to rewrite this as the limit, as t approaches infinity
of negative 54 over e raised to the power
of positive 1/360 times t plus 54.
Notice as t approaches infinity,
this denominator here increases without bound,
and therefore this term approaches zero,
and therefore this limit is equal to zero plus 54,
which equals 54.
So this tells us, as time increases without bound,
the amount of sugar in the tank approaches 54 kilograms.
I hope you found this helpful.