Author Topic: Check my work please?  (Read 3509 times)

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Offline while1

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Check my work please?
« on: April 05, 2007, 11:17:29 pm »
Alright, the problem is:



Alright.  Here's what my solution/answer came out to be:



Is this correct?  Methinks I should take DiffEq even though it's not a major req.
« Last Edit: April 06, 2007, 02:45:56 am by Michael »
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Offline dark_drake

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Re: Check my work please?
« Reply #1 on: April 06, 2007, 12:21:41 am »
That seems really wrong to me.

The first thing that strikes me as odd is your dx/dt equation. The rate in looks right, but the rate out seems a bit odd.  Why is there (600+t)gal on the bottom?  There should be 600, but why do you have the +t? It makes absolutely no sense to me; the amount of water isn't changing with time. I think the rate out should be x lb/600 gal * 5 gal/min.

And then what you have is dx/dt= (600-x)/120.

dx/(600-x)=dt/120.

Integrate both sides for: -ln|600-x| = t/120 +C.

In the end, I get 567.2 lbs. after 5 hours.

Edit: NOOOOOO! This was my 1337th post, and I didn't even notice it!!
« Last Edit: April 06, 2007, 01:04:45 am by dark_drake »
errr... something like that...

Offline while1

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Re: Check my work please?
« Reply #2 on: April 06, 2007, 12:51:43 am »
That seems really wrong to me.

The first thing that strikes me as odd is your dx/dt equation. The rate in looks right, but the rate out seems a bit odd.  Why is there (600+t)gal on the bottom?  There should be 600, but why do you have the +t? It makes absolutely no sense to me; the amount of water isn't changing with time. I think the rate out should be x lb/600 gal * 5 gal/min.

And then what you have is dx/dt= (600-x)/120.

dx/(600-x)=dt/120.

Integrate both sides for: -ln|600-x| = t/120 +C.

In the end, I get 567.2 lbs. after 5 hours.
hmmm.  The reason why I have the (600+t) is because the volume of salt water in the vat (in gallons) isn't constant (doesn't stay at 600 gallons of salt water).  So if x represents the amount of salt in the vat then I was thinking that the volume of the vat at time t = (600+t)... so x/(600+t) gives the proportion of salt to the salt water in the vat at time t.  Not sure if my reasoning is right though.  Someone else want to give their opinion in support of either of us or neither (we're both wrong).
« Last Edit: April 06, 2007, 12:59:00 am by Michael »
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Offline dark_drake

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Re: Check my work please?
« Reply #3 on: April 06, 2007, 12:55:42 am »
hmmm.  The reason why I have the (600+t) is because the volume of salt water in the vat (in gallons) isn't constant (doesn't stay at 600).  So if x being the amount of salt in the vat and (600+t) = volume of salt water in vat.  So x/(600+t) would give the proportion of salt in the vat.  Not sure though.
The volume of solution will stay constant! It is constantly mixed, so the concentration of the salt in the water will always be x/600. And concentration of solution * rate of solution leaving = rate out = 5x/600.
« Last Edit: April 06, 2007, 01:02:24 am by dark_drake »
errr... something like that...

Offline dark_drake

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Re: Check my work please?
« Reply #4 on: April 06, 2007, 01:00:14 am »
Another useful check for problems like these is that that it the rates should eventually reach an equillibrium; yours never does. There should come a point in time where the rate of salt entering = the rate of salt leaving, and that will be as high as the level of solute will get. You do not have that in your final equation. As time goes to infinity in your equation, so does your salt levels.

With mine, however, if you solve it comes to x=600-400*e^(-t/120). The limit of x as t approaches infinity is 600; this is where rate in = rate out.
« Last Edit: April 06, 2007, 02:48:54 am by dark_drake »
errr... something like that...

Offline while1

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Re: Check my work please?
« Reply #5 on: April 06, 2007, 01:04:44 am »
hmmm.  The reason why I have the (600+t) is because the volume of salt water in the vat (in gallons) isn't constant (doesn't stay at 600).  So if x being the amount of salt in the vat and (600+t) = volume of salt water in vat.  So x/(600+t) would give the proportion of salt in the vat.  Not sure though.
The volume of solution will stay constant! It is constantly mixed, so the concentration of the salt in the water will always be x/600. And concentration of solution * rate of solution leaving = rate out = 5x/600.
k, now I understand.  The concentration part was what I didn't grasp... referring to it as 'concentration' is what made it click in my head/make me think 'duh'.  Thanks.
« Last Edit: April 06, 2007, 01:09:05 am by Michael »
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Offline dark_drake

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Re: Check my work please?
« Reply #6 on: April 06, 2007, 02:54:48 am »
Glad to be of help.

A helpful trick is to check if your x(t) equation makes sense is check what the limit as time approaches infinity is. Provided you set up your dx/dt equation correctly, the final concentration of salt should be the same as solving rate in = rate out.

For example, with this problem, I got:

Rate in = 5 lbs./min.

Rate out = x lbs./120 min.

5 = x/120

x=600.

And this matces with the limit as time approaches infinity of my x(t).

I hope that makes sense.
errr... something like that...