Diff Eq. - Rate In/Out Question (Heeeellllppp!)

[mooeypoo]

Hey,

 

I have a question in the book, solved it one way, and the book solution in cramster has another way. I am not sure if I'm right or not, since in the case of these questions, multiple cases for solutions are possible.

 

Any help would be appreciated here.. where was I wrong (was I?)

 

Question:

A Tank Contains 100 gallons of water and 50 oz of salt. Water containing a salt concentration of

[math]\dfrac{1}{4} (1+\frac{1}{2} sint)[/math] oz/gal flows into the tank at a rate of 2 gal/min, and the mixture in the tank flows out at the same rate.

(a) Find the amount of salt in the tank at any time.

So.. I started with the premise that

[math]\frac{dQ}{dt} = Rate In - Rate Out[/math]

 

Rate In:

[math]\frac{2}{4} (1+ \frac{1}{2} sint)Q[/math]

 

Rate Out:

[math]Q/100 * 2[/math]

 

So,

 

[math]\frac{dQ}{dt} = \frac{1}{2} (1+ \frac{1}{2} sint)Q - Q/50[/math]

 

[math]= Q\left( \dfrac{1}{2} (1+2sint)-\frac{1}{50} \right) [/math]

 

[math] \frac{1}{Q} dQ - \left( \frac{1}{2} (1+2sint)+\frac{1}{100} \right) dt = 0[/math]

 

[math]\int \frac{1}{Q} dQ - \int \frac{1}{2} (1+2sint) dt + \int \frac{1}{100} dt = 0[/math]

 

[math]lnQ - \frac{t}{2} - cost + \frac{t}{100} = C[/math]

 

[math]lnQ - cost - \frac{t}{100} + \frac{50t}{100} = C[/math]

 

[math]lnQ = cost - \frac{49t}{100} + C[/math]

 

[math]Q = C \exp{( -\frac{49t}{100}+cost)}[/math]

 

Then, I can test for C by putting the initial values Q=50 when t=0:

 

[math]50 = C \exp{(cos0)} = Ce[/math]

 

[math]C = \frac{50}{e}[/math]

 

And, therefore, the final equation is:

 

[math]Q = 50 \exp{(\frac{49t}{100}+cost-1)}[/math]

 

But the answer in cramster.com (I don't have an answer in my book), says:

 

[math]

Q = 25 - \frac{625}{2501} cost + \frac{25}{5002} sint + \frac{63150}{2501} \exp{(-\frac{t}{50})}

[/math]

 

Thanks in advance, this is my preparation for a test, so I'd love to know where I got things wrong.. I was actually quite happy to see my dQ/dt equation is separable, but cramster.com seems to not even try the separable equation technique and goes on straight to integrating factor.

 

Should I go straight onto integrating factors without even checking if the equation is separable? Is that a better method?

 

Anyhoo, I'd appreciate an answer.. the exam is on monday and I want to make sure I am at laest preparing correctly...

 

Thanks!

 

~moo

Edited October 3, 2008 by mooeypoo

[NeonBlack]

Hey,

Rate In:

[math]\frac{2}{4} (1+ \frac{1}{2} sint)Q[/math]

 

Is this part correct?

Does the rate in have anything to do with the amount of salt already in the tank?

[insane_alien]

the amount of salt in the tank will vary with time. you need the rate in and yes it is correct as it has a flowrate of 2 gal/hour.

[NeonBlack]

i_a, you have misunderstood.

Rate In:

[math]\frac{2}{4} (1+ \frac{1}{2} sint)Q[/math]

 

Here is a better way to write that:

[math]\frac{dQ_{in}}{dt}=2\frac{gal}{min}\frac{1}{4} (1+ \frac{1}{2} sint)\frac{oz}{gal}Q (oz)[/math]

 

Do you see a problem now?

This is why I prefer symbolic solutions. Well, that and I'm not very good at arithmetic.

Edited October 3, 2008 by NeonBlack

[mooeypoo]

Uh, no no..

 

Rate In:

[math]\frac{1}{4} (1+ \frac{1}{2} sint)Q \frac{(oz)}{(gal)} * 2 \frac{(gal)}{(min)} = \frac{2}{4} (1+ \frac{1}{2} sint)Q \frac{(oz)}{(min)}[/math]

 

Rate Out:

[math]\frac{Q (oz)}{100 (gal)} * 2 \frac{(gal)}{(min)} = \frac{2Q (oz)}{100 (min)} = \frac{Q (oz)}{50 (min)}[/math]

 

So,

 

[math]\frac{dQ}{dt} = \frac{1}{2} (1+ \frac{1}{2} sint)Q \frac{(oz)}{(min)} - \frac{Q (oz)}{50 (min)}[/math]

 

It's kinda hard writing things in LaTeX, but it does work out (go over it again .

 

This part of the question is identically solved in my notebook and cramster.com , btw.. the problem starts with SOLVING this equation.

[NeonBlack]

I'm sorry. First, you said:

A Tank Contains 100 gallons of water and 50 oz of salt. Water containing a salt concentration of [math]

\dfrac{1}{4} (1+\frac{1}{2} sint)

[/math] oz/gal flows into the tank at a rate of 2 gal/min, and the mixture in the tank flows out at the same rate.

[mooeypoo]

I'm sorry. First, you said:

Uh, yes.. and what I wrote as "Rate In" and "Rate Out" was the representation of that... I don't see the problem.. did I get some LaTeX wrong or something?

[NeonBlack]

I was trying to be subtle so that you would see it for yourself.

What you wrote as "rate in" has an additional factor of Q.

This is what I am saying: (taking things directly from the problem)

[math]

\dfrac{1}{4} (1+\frac{1}{2} sint) oz/gal \times 2 gal/min = \dfrac{1}{2}(1+\frac{1}{2} sint) oz/min

[/math]

Since (you like chemistry, don't you?) amount=concentration x volume.

[mooeypoo]

I was trying to be subtle so that you would see it for yourself.

What you wrote as "rate in" has an additional factor of Q.

This is what I am saying: (taking things directly from the problem)

 

Okay, for the past 3 posts I am not getting it, and I don't get it still.

 

Can you please show where the extra factor of Q is, please? The question states that the is the concentration. Concentration of salt. Hence it needs to be written as Q

 

Is that what you mean?

 

You can think of the formula, instead, as if it says 43%.

43% of what? of Q. So the rate IN is 47/100 OF Q ==> 47/100*Q

 

Same here, only with sint and fractions.

 

 

If I wrote the equation wrong, please point out where, this is getting quite frustrating.

[NeonBlack]

Look at the units it gives you for the concentration:

oz(of salt)/gal(of water)

Here, there is no "of what?" question, simply because it tells you. Something like 43% is dimensionless.

[mooeypoo]

but but.. but..

 

 

Okay, I was about to post a whole mumbly confused question about this, but while I was typing, I suddenly got it...

 

My professor is making the mistake of telling us to NOT write units at all because they are cancelling out anyways. It's a math prof, not physics or chem (btw, i hate chem, but that's a side note ) so i guess it's working out for him. I always write units, not just to get the final units but to also see where I get things wrong.

 

Like here.

 

I am not sure why, maybe because I read a few of his explanations (lacking units), but this time I wrote the units and actually missed the concentration units.

 

Damn.

 

At least I have a case to show him and prove my point (I actually argued this in class )

 

Thanks I'll try it again now and see if I can solve the equation when it's actually written correctly.

 

Now as a side note -- *IF* this was the right equation, I could do it as separable variables method, right? of course integrating factor method would work here too, I guess, but is there any reason for me to pick one and not the other when solving such equations, or is it a matter of pure preference when the eq. can be solved using different methods?

 

Thanks again

 

~moo

[NeonBlack]

No, separation of variables won't work, unfortunately, because there is a term 1/2+1/4sin(t).

You can do integrating factor, but (in my opinion) it's easier to do undetermined coefficients, letting Qparticular=Asint+Acost+C

[mooeypoo]

No, separation of variables won't work, unfortunately, because there is a term 1/2+1/4sin(t).

You can do integrating factor, but (in my opinion) it's easier to do undetermined coefficients, letting Qparticular=Asint+Acost+C

 

Oh, no, I know that it won't work in this case, I was refering to general cases where I can use both -- is there any preference?

 

I .. am not sure we learned undetermined coefficients. I will definitely check it out..