February 29th, 2012, 05:20 AM  #1 
Member Joined: Sep 2011 Posts: 97 Thanks: 0  expression in mixing problem
A tank contains 100 gal of fresh water. A solution containing 1 lb/gal of soluble lawn fertilizer runs into the tank at the rate of 1 gal/min, and the mixture is pumped out of the tank at the rate of 3 gal/min. Find the amount of fertilizer F in the tank as a function of time t. what i did was this 100gal of water 0lb lawn in the initial phase 1 gal of water with 1 lb/min input 3 gal of water output rate out: rate in: then the equation next i going to get the integrating factor i'm stucked here, because i don't know how to integrate that 
February 29th, 2012, 11:20 AM  #2 
Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 520 Math Focus: Calculus/ODEs  Re: expression in mixing problem
Let represent the amount in lbs. of fertilizer in the tank at time t. Let represent the volume of the mixture at time t, where V is measured in gallons and t is measured in minutes. We have then: where So we see that in 50 minutes, the tank will be empty. Now we may state: where and We then calculate the integrating factor as: Here is a plot: [attachment=0:qoa7vp9r]fertplot.jpg[/attachment:qoa7vp9r] 
February 29th, 2012, 05:27 PM  #3 
Member Joined: Sep 2011 Posts: 97 Thanks: 0  Re: expression in mixing problem
there are some equations manipulation that i don't understand, and my lecturer and the lecture notes been skipping those parts too how did this become to this then to this 
February 29th, 2012, 05:35 PM  #4 
Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 520 Math Focus: Calculus/ODEs  Re: expression in mixing problem
Let's begin with: A bit of factoring on the second term on the left side lets us write: Which we may rearrange as: and so we now have: Now, the whole point to getting the proper integrating factor is so that we may express the left side as the differentiation of a product. Observe that: Hence we may now write the ODE as: 
February 29th, 2012, 06:26 PM  #5 
Member Joined: Sep 2011 Posts: 97 Thanks: 0  Re: expression in mixing problem this part i don't understand, sorry... try to think deeper but still don't know how you conclude the right side is equal to the left side is it because you throw away the but keep the A? 
February 29th, 2012, 06:29 PM  #6 
Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 520 Math Focus: Calculus/ODEs  Re: expression in mixing problem
It is an application of the product rule, since A is a function of t. Apply the product rule to the left side and you will get the right side.

February 29th, 2012, 06:32 PM  #7 
Member Joined: Sep 2011 Posts: 97 Thanks: 0  Re: expression in mixing problem
oh ok, i suppose it's this law right? h'(x) = f'(x)g(x) + f(x)g'(x) 
February 29th, 2012, 06:36 PM  #8 
Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 520 Math Focus: Calculus/ODEs  Re: expression in mixing problem
Yes, more commonly it is written in the order: or using prime notation: Use this rule and you should get the above conclusion. 
February 29th, 2012, 07:51 PM  #9 
Member Joined: Sep 2011 Posts: 97 Thanks: 0  Re: expression in mixing problem
just one more tiny question then what is g(x) 
February 29th, 2012, 08:01 PM  #10 
Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 520 Math Focus: Calculus/ODEs  Re: expression in mixing problem 

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