Differential Equations

After a long hiatus from my last college math class (which was DifEq, which I did somehow pass), I’ve decided to revisit it. I’m starting with the classic problem of:

There’s a full 100 liter tank containing a water and 10kg salt. If we add 10 L pure water per minute while simultaneously draining 10 L of the solution per minute, how much salt will be left in the tank after 30 minutes?

I’m having a hard time setting up the formulas. Does anybody have an intuitive understanding of DifEq?

DEs have always been one of my weak spots.

However, this site seems to have the solution you desire.

I was about to say, "Please tell me I didn't just do someone's homework for him," but given that the link is to a university website, if a university wants to make the answer public...

Quote: Ace2

...I’m having a hard time setting up the formulas...

In class, my engineering professor started every example problem like yours with this very useful equality:
Input - Output = Accumulation

He would then substitute expressions for each of the three terms. For your problem, the expressions would be:
Input of salt = 0 x 10 L/m = 0
Output of salt = concentration x 10 L/m = 10f, where f is the varying concentration in kg salt/l of water
Accumulation of salt = d(100f)/dt = 100 df/dt

So, the differential equation to solve is:
-10f = 100 df/dt

After solving this, put in the initial condition:
f(0) = 10/100

Then find f(30) and multiply by 100 to find the salt remaining in the tank after 30 minutes.

Quote: Ace2

There’s a full 100 liter tank containing a water and 10kg salt. If we add 10 L pure water per minute while simultaneously draining 10 L of the solution per minute, how much salt will be left in the tank after 30 minutes?

Let:
s = kg of salt.
t = minutes since salt was dumped into the tank.

We're given:

ds/dt = -(10/100) * s
ds = (-10/100)*s dt
-10/s ds = dt

Take the integral of both sides.

-10*ln(s) = t + c

Next, let's solve for the dreaded constant of integration. We're given that at t=0, s=10. So...

-10*ln(10) = 0 + c
c = -10*ln(10)

So, our equation of salinity is now:

-10*ln(s) = t - 10*ln(10)

The question at hand is how much salt will be in the tank at t=30.

-10*ln(s) = 30 - 10*ln(10)

ln(s) = -3 + ln(10)

s = exp(-3 + ln(10))
s = exp(-3) * exp(ln(10))
s = 10 * exp(-3)
s = 0.4979 kg of salt.

I was hoping to hear an "a men" from one of the other math geeks of the forum.

ok... Amen!

Quote: rsactuary

ok... Amen!

Thanks. I may have no clue how to talk with women, but I can at least remember how to do an ordinary differential equation. At least I think I remember.

Thanks for the replies.

I was busy shooting dice in Vegas over the long weekend but now I’ll have time to review the answers in detail.

I see the actual formulas are quite basic so this is just a matter of me studying the setup/answers until the light bulb turns on. I’ve always been a bit slow when it comes to anything related to calculus.