- Welcome to an application problem
involving a separable differential equation.
Here we have a mixture problem.
A tank contains 90 kilograms of salt
and 2000 liters of water.
Pure water enters a tank
at a rate of eight liters per minute.
The solution is mixed and drains from the tank
at a rate of four liters per minute.
The first question is:
What is the amount of salt in the tank initially?
Well, the first sentence tells us
we're starting with 90 kilograms of salt.
So part a is 90 kilograms.
This also gives us what's called the initial condition,
where if the function of y of t
tells us how much salt is in the tank at any time t,
we now know that y of zero must equal 90.
We'll need this information again later.
Part b, we're asked to find the amount of salt
in the tank after one point five hours,
and then c, I'm going to find the concentration of salt
in the solution in the tank as time approaches infinity.
We can model the situation using the differential equation
dy, dt equals R sub one minus R sub two,
where dy, dt is the rate of change
of the amount of salt at any time t.
R sub one is the rate of change
of the amount of salt entering the tank,
and R sub two is the rate of change
of the amount of salt exiting or leaving the tank.
So we'll begin by finding R sub one and R sub two.
For these rates, we'll find the product of the concentration
of the solution and the flow rate.
Again, R sub one is the rate of change of the salt
entering the tank, so we'll have the concentration
of the salt in the solution entering the tank,
times the flow rate.
Notice how we're told pure water enters a tank
at a rate of eight liters per minute.
So the concentration of salt
in the solution entering the tank
is actually zero kilograms of salt per liter,
because it's pure water,
times a flow rate of eight liters per minute.
Notice how the units of liters does simplify out,
but either way the product is zero.
It would be zero kilograms of salt
entering the tank per minute.
And now for R sub two we want to find the rate
at which salt is entering the tank.
So again we'll have a concentration
of salt times a flow rate,
and we're told the flow rate exiting the tank,
or leaving the tank, is four liters per minute.
Now the concentration of salt in the tank
at any time t is actually changing.
It's based upon the amount of salt in the tank at any time t
and the amount of water in the tank at any time t.
We know the amount of salt in the tank at any time t
is equal to y, or y of t.
So we'd have y divided by the total amount of water
in the tank at any time t,
is going to be 2000, the starting amount,
plus notice how the flow rate entering the tank
is eight liters per minute,
and the flow rate exiting the tank
is only four liters per minute.
Which means the amount of water in the tank is increasing
by four liters per minute,
because the flow rate entering is four liters more
than the flow rate exiting.
So we have 2000 plus four t,
for the total amount of water in the tank at any time t.
And the units here are kilograms per liter.
Liters simplifies out, and this gives us R sub two,
equals four y divided by the quantity 2000 plus four t,
and the units are kilograms per minute.
But notice how this does simplify,
all the terms contain a factor of four.
This simplifies to y
divided by the quantity 500 plus t kilograms per minute.
Which means our differential equation is going to be dy, dt,
equals R sub one, which is zero, minus R sub two,
which again is y divided by the quantity 500 plus t.
Let's go ahead and solve this on the next slide.
So we'd have dy, dt equals, we can drop the zero
and just write negative y, all over the quantity 500 plus t.
Again this is a separable differential equation
if we multiply both sides by dt,
we have differential y equals negative y
divided by the quantity 500 plus t, dt.
We want the y term to be on the right side,
let's go ahead and multiply both sides by one over y.
So we'd have one over y, dy,
equals negative one over the quantity 500 plus t, dt.
So now that we have a function of y times dy,
equals a function of t times dt.
We integrate both sides of the equation,
integrate, and then solve for y.
So we'd have the integral of one over y, dy
equals the integral of negative one over 500 plus t, dt.
So on the left side we have natural log,
absolute value of y.
We would have a constant of integration,
but we also have a constant on the right,
so lets only but the constant on the right.
So we'd have equals.
Now you might be thinking
we have to perform u substitution here,
notice how if we let u be equal to 500 plus t,
differential u is equal to one dt, or just dt,
so no u substitution is required.
On the right side we'll just have negative natural log,
absolute value of the quantity 500 plus t.
And for the constant of integration,
let's write plus C sub one.
Now we need to solve this for y.
To do this though, we'll need to get the natural logs
on the same side.
Let's add natural log of the quantity
500 plus t to both sides,
which would give us natural log absolute value of y,
plus natural log absolute value of the quantity 500 plus t,
equals C sub one.
Now here, because we're adding two logs with the same base,
we can combine them by multiplying y and 500 plus t.
So we can say natural log, absolute value of y
times a quantity 500 plus t, equals C sub one.
Now lets write the log equation as an exponential equation
where we know that natural log is log base e,
so as an exponential equation
we would have e raised to the power of C sub one,
equals y times the quantity 500 plus t.
So let's write that as y times the quantity 500 plus t,
equals e to the power of C sub one.
But e raised to the power of C sub one
is just some constant,
let's have e raised to the power of C sub one be equal to C.
So we'd have y times the quantity 500 plus t, equals C.
Solving for y, we'd divide both sides by 500 plus t.
So now we know that y, or y of t,
equals the constant C divided by the quantity 500 plus t.
To find the value of C,
we use the initial condition that we found on part a,
that y of zero equals 90.
So if y of zero equals 90, we can substitute 90 for y of t,
and zero for t and solve for C.
That would give us the equation 90 equals C
divided by the quantity 500 plus zero, or just 500.
So multiplying both sides by 500 we'd have C equals 45000.
By substituting 45000 for C,
we now have the particular solution for the function y of t
that will tell us how much salt is in the tank
at any time t.
We would have y of t equals 45000
divided by the quantity 500 plus t.
Going back to our question,
part b asked us to find the amount of salt in the tank
after one point five hours.
Really be careful here,
because the time in our problem is in minutes, not hours.
So to answer this question,
because one point five hours is equal to 90 minutes,
we want to find y of 90.
Using our function y of t,
we want to determine the value of y of 90.
So we'd have 45000 divided by quantity 500 plus 90,
which is 45000 divided by 590.
Which would be approximately 76.2712.
And the units here would be kilograms of salt.
So going back to our first slide,
we now know part b is 76.2712 kilograms of salt.
For part c we're asked to find the concentration of salt
in the solution in the tank as the time approaches infinity.
If we find the limit as t approaches infinity of y of t,
that would tell us the amount of salt.
Once you find the amount we can find the concentration.
Let's have the limit as t approaches infinity of y of t.
Would actually be equal to zero,
because notice how the denominator increases without bound
because t approaches infinity,
and the numerator is a constant,
which means as time increases without bound,
the amount of salt in the tank approaches zero,
and therefore the concentration would also approach zero.
So part c, the concentration approaches zero
kilograms of salt per liter.
I hope you've found this helpful.