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**Reuel****Member**- Registered: 2010-11-28
- Posts: 178

Hello.

I am still getting the hang of mixture problems. Here is one.

"A mixing tank initially contains 120 gallons of brine which in turn contains 75 pounds of salt in its solution. A new brine containing 1.2 pounds of salt per gallon begins entering the tank at t=0 and at the rate of 2 gallons per minute while the uniform mixture flows out of the tank at 1 gallon per minute. Assuming the mixture is kept always uniform, find the amount of salt in the tank at the end of 1 hour."

And here is how I have it set up:

I would like some feed back on how it is set up because, done this way, it is not separable and therefore needs one of those exponentials with an integral in the power.

How's it look? And yep, I am solving it by hand.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 105,702

Hi;

I am not getting your answer, why?

I get:

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.****No great discovery was ever made without a bold guess. **

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**Reuel****Member**- Registered: 2010-11-28
- Posts: 178

The mix is leaving the tank slower than it is coming in.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 105,702

But is not the amount of water at time t,

w(t) = 120 + 1*t = 120 + t ?

( 2 gals in - 1 gal out ) = 1 gal.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.****No great discovery was ever made without a bold guess. **

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**Reuel****Member**- Registered: 2010-11-28
- Posts: 178

Sorry.... went to bed and had other things going on today.

So... you are correct. I realize why it would be only t and not 2t. So here is my work. Is my solution satisfactory? (I did not show too many steps this time.)

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**Reuel****Member**- Registered: 2010-11-28
- Posts: 178

I am confident in that solution but I am also very tired so if I made a mistake I'll fix it!

I am about to post another more difficult problem. I hope that all these solutions help others looking for worked out problems like these!

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 105,702

That is the answer I am looking for, 170. That is correct as far as I know. Good work!

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.****No great discovery was ever made without a bold guess. **

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**Reuel****Member**- Registered: 2010-11-28
- Posts: 178

Here is another mixture problem and my solution. I hope it is correct, for all the problems I have done by now.

"A tank starts out containing 50 gallons of brine which holds 30 pounds of salt within its solution. Water begins to pour into the tank at 3 gallons per minute and the well-stirred solution runs out of the tank at 2 gallons per minute. How long before there will be 25 pounds of salt in the tank?"

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 105,702

Hi;

Looks good!

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.****No great discovery was ever made without a bold guess. **

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**algebzz****Guest**

Reuel wrote:

Sorry.... went to bed and had other things going on today.

So... you are correct. I realize why it would be only t and not 2t. So here is my work. Is my solution satisfactory? (I did not show too many steps this time.)

how did you get this answer? can you explain it to me..

I really appreciate f you do..

**kentoi_tisoy12****Member**- Registered: 2011-02-20
- Posts: 1

to reuel, I think I kinda agree with the request of algebzz..

Im kinda confuse. Would you please show me how you raise with e the integral of dt/120 + t?

Thanks. Is it a first ordered linear equation?

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**Reuel****Member**- Registered: 2010-11-28
- Posts: 178

I'm so sorry - I didn't see your questions from a couple of months ago.

The first-order differential equation is not separable because it is a subtraction problem and not one of multiplication or division - see how it has a minus sign. And so we need what is called an integrating factor. Here are the steps:

1) Move the part of the equation on the right hand side with the dependent variable in it to the left hand side.

2) Compute the integrating factor. That is, make up an equation where e is raised to the power of the integral of the dependent variable's maths without the independent variable itself. Here is a youtube video that teaches about the integrating factor:

http://www.youtube.com/watch?v=Et4Y41ZNyao

3) Once you have the integrating factor you can plug it in as shown in line 4 of my work you quoted.

Sorry again for not getting back to you. I am preparing for my final and am reviewing my old posts and just happened to see your questions. I hope this has helped you out. I can provide a few of my own examples with steps if you would like!

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