Differential Equations: Compartment Model, when I = 0

[royblaze]

Homework Statement

The pollutants are being added to the lake at a constant rate I, and the pollutants are thoroughly mixed into the lake.

The annual precipitation into the lake matches evaporation, so the flow rate F is also constant.

Let y(t) be the amount of pollutant in the lake and c(t) its concentration at time t.

Suppose that, after reaching the long term concentration, the pollution of the lakes is stopped. Then I = 0 and the differential equation reduces to c' = -(F/V)c. But this is the decay version of the Malthusian model (i.e. exponential decay) whose solution is given by c(t) = c₀ e^(F/V)t.

In the case of Lake Erie, V = 458 cm³ and F = 175 km³/year. How many years will it take for c(t) to drop from c_infinity to 1/10 c_infinity?

Homework Equations

The differential equation for y in this one-compartment model is y' = - (F/V)y + I. Simply dividing this by V and recalling that c(t) = y(t)/V, we find that the differential for c is:

c' = -(F/V)c + I/V

The Attempt at a Solution

I tried taking those given F and V values and inputting them into the equation. However, the question asks how many years it would take to change c_infinity. But if you take the limit of c(t) as t approaches infinity, then wouldn't the function c(t) just equal infinity? How can I go about setting some "infinity" value equal to one-tenth that "infinity"? I'm so confused! Please help!

 

[Kreizhn]

Hmm...you seem to have presented the question in a very backwards sense. I'm reading your problem statement and the definition of each term appears below it.

First of all, is I=0 in your model? Secondly, notice that the sign of your exponential is wrong. It should be
c(t) = c_0 e^{-\frac FV t}
so that when t \to \infty, c(t) doesn't explode.

 

[royblaze]

Hum, well on the question it states that the c(t) = c₀e^(F/V)t.

How does one get the (-) negative sign in front of the F/V ? And yes, the I = 0 in the model.

So if the c(t) doesn't explode into infinity then, doesn't it just go to zero?

 

[Kreizhn]

Well, if the differential equation is
\frac{dc}{dt}(t) = -\frac FV c(t)
then the negative sign definitely needs to appear. Try solving this by separating the variables. That is, rewrite it at
\frac{dc}{c} = -\frac FV dt
and then integrating, remembering your constant of integration.

You are correct, if I=0 then c(t) \xrightarrow{t\to\infty} 0 [/tex]. The interesting things happen when I \neq 0 because then this is no longer true.

 

[royblaze]

Okay, so I have now:

c = e^(-F/V)t + e^k

Where k is the constant (both sides have a constant, but I just combined them into a general constant). I also can see that e to a constant is just a constant as well, but I left it in there.

So I see where you got the negative. But then how do I actually go about solving the question? Taking this limit as it approaches infinity results in zero, like you said. But then if that limit equals zero, then how does ones get 1/10 of c_infinity?

 

[Kreizhn]

Be careful with your constants!

In particular, you should have gotten

\log(c) =-\frac FV t + k
right? So when you exponentiate, the sum becomes mulitplication:

c(t) = e^{-\frac FV t + k} = e^{-\frac FV t}\underbrace{ e^k}_{c_0}

As for the issue with c_\infty, you're right, it doesn't make any sense. It seems much more likely that the question wants you to consider the problem when I \neq 0, since in this case c_\infty \neq 0. Maybe double-check the question and see if that's what they're asking?

 

[royblaze]

Oh hot darn, I can't believe I fuddled that part up.

So far in my class, we've been doing compartment models where we actually have a numerical value for I. This is the first question so far where we're evaluating a situation without the I. I'll post an image of the full question, for the sake of the question. Maybe someone can interpret it better than I can.

 

[Kreizhn]

Hmm...I don't like how the question states it, but let's give it a shot.

First of all, do you know the answer to question 4?

Secondly, I think we can make the assumption that there are typo's in the question given that the sign of the exponential was wrong. What I think the question means to ask is

"How long will it take for c(t) to drop from c_0 to \frac1{10} c_0. "

 

[royblaze]

Yes, I answered question four. The answer is D.

 

[Kreizhn]

Good. It's not important but it's good to see you know it. Now try answering the question with c_0 instead of c_\infty.

 

[royblaze]

Answering question 4 or 5 with the c₀?

My last line before taking a limit (from question 4) is

c(t) = I/F + Ce^-(F/V)t

I'm still confused about how we're solving it. "Now try answering the question with c₀" ? Sorry I'm not catching on too quick.

 

[Kreizhn]

Don't worry about it.

Forget question 4.

We're now working with question 5, in which case I=0, and so we computed the solution to be

c(t) = c_0 e^{-\frac FV t}

So I think the question should read

"What value of t gives c(t) = \frac1{10} c_0?"

Now I've solved this question and the answer is indeed listed. I think that this is what you're supposed to do.

 

[royblaze]

Thanks for helping me out so far along. I've been trying to riddle this out for the last 20 minutes, though. I tried a bunch of ways, but so far this is the most recent one, where I got a legitimate number...

If we're trying to find a t to make c(t) = 1/10 c₀, then what I did was solve for t after inputting the values.

if c(t) = 1/10 c₀, and c(t) = c₀ * e^-(F/V)t, then is it "legal" to say that:

e^(-F/V)t = 1/10 ?

I solved for t in that case and found ln(10) * 458...

 

[royblaze]

I worked on it a little longer, I got an answer of around 6. Hope that's what you got. Thanks for the help!

 

[Kreizhn]

Yeah, I got 6.02 or something to that effect.

Edit: Sorry that I didn't reply to your last post. I got a flourish of PF emails and must have missed the fact that you replied.

 

[royblaze]

Nah, I appreciate the help. Thanks again!