My Math Forum expression in mixing problem

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 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 $V(t) = 100 gal + (1 gal/min -3 gal/min)(t min) V(t) = 100 - 2t$ rate out: $[y/(100 - 2t)] (3)= 3y/(100-2t) lb/min$ rate in: $(1 gal * 1 lb)= 1 lb/min$ then the equation $dy/dt = 1 - 3y/(100-2t) dy/dt + 3y/(100-2t) = 1$ next i going to get the integrating factor $e ^ (|P dt)= e ^(| (3y/(100-2t)))$ i'm stucked here, because i don't know how to integrate that
February 29th, 2012, 11:20 AM   #2
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Re: expression in mixing problem

Let $A(t)$ represent the amount in lbs. of fertilizer in the tank at time t. Let $V(t)$ represent the volume of the mixture at time t, where V is measured in gallons and t is measured in minutes. We have then:

$\frac{dV}{dt}=1-3=-2$ where $V_0=100$

$\int_{V_0=100}\,^V\,dV=-2\int_{0}\,^t\,dt$

$V-100=-2(t-0)$

$V=100-2t$

So we see that in 50 minutes, the tank will be empty.

Now we may state:

$\frac{dA}{dt}=$$1\text{ \frac{gal}{min}}$$$$1\text{ \frac{lb}{gal}}$$-$$3\text{ \frac{gal}{min}}$$$$\frac{A(t)}{100-2t}\:\text{\frac{lb}{gal}}$$$

$\frac{dA}{dt}+\frac{3}{100-2t}A=1$ where $A_0=0$ and $0\le t<50$

We then calculate the integrating factor as:

$\mu(x)=e^{\int\frac{3}{100-2t}\,dt}=(50-t)^{\small{-\frac{3}{2}}}$

$(50-t)^{\small{-\frac{3}{2}}}\frac{dA}{dt}+\frac{3}{100-2t}(50-t)^{\small{-\frac{3}{2}}}A=(50-t)^{\small{-\frac{3}{2}}}$

$(50-t)^{\small{-\frac{3}{2}}}\frac{dA}{dt}+\frac{3}{2}(50-t)^{\small{-\frac{5}{2}}}A=(50-t)^{\small{-\frac{3}{2}}}$

$\frac{d}{dt}$$(50-t)^{\small{-\frac{3}{2}}}A$$=(50-t)^{\small{-\frac{3}{2}}}$

$\int\,d$$(50-t)^{\small{-\frac{3}{2}}}A$$=\int(50-t)^{\small{-\frac{3}{2}}}\,dt$

$(50-t)^{\small{-\frac{3}{2}}}A=2(50-t)^{\small{-\frac{1}{2}}}+C$

$A(t)=2(50-t)+C(50-t)^{\small{\frac{3}{2}}}$

$A(0)=2(50-0)+C(50-0)^{\small{\frac{3}{2}}}=0$

$C=-\frac{\sqrt{2}}{5}$

$A(t)=2(50-t)-\frac{\sqrt{2}}{5}(50-t)^{\small{\frac{3}{2}}}$

Here is a plot:

[attachment=0:qoa7vp9r]fertplot.jpg[/attachment:qoa7vp9r]
Attached Images
 fertplot.jpg (10.3 KB, 113 views)

 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 $(50-t)^{\small{-\frac{3}{2}}}\frac{dA}{dt}+\frac{3}{100-2t}(50-t)^{\small{-\frac{3}{2}}}A=(50-t)^{\small{-\frac{3}{2}}}$ become to this $(50-t)^{\small{-\frac{3}{2}}}\frac{dA}{dt}+\frac{3}{2}(50-t)^{\small{-\frac{5}{2}}}A=(50-t)^{\small{-\frac{3}{2}}}$ then to this $\frac{d}{dt}$$(50-t)^{\small{-\frac{3}{2}}}A$$=(50-t)^{\small{-\frac{3}{2}}}$
 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: $(50-t)^{\small{-\frac{3}{2}}}\frac{dA}{dt}+\frac{3}{100-2t}(50-t)^{\small{-\frac{3}{2}}}A=(50-t)^{\small{-\frac{3}{2}}}$ A bit of factoring on the second term on the left side lets us write: $(50-t)^{\small{-\frac{3}{2}}}\frac{dA}{dt}+\frac{3}{2(50-t)}(50-t)^{\small{-\frac{3}{2}}}A=(50-t)^{\small{-\frac{3}{2}}}$ Which we may rearrange as: $(50-t)^{\small{-\frac{3}{2}}}\frac{dA}{dt}+\frac{3}{2}\cdot\frac{( 50-t)^{\small{-\frac{3}{2}}}}{50-t}A=(50-t)^{\small{-\frac{3}{2}}}$ and so we now have: $(50-t)^{\small{-\frac{3}{2}}}\frac{dA}{dt}+\frac{3}{2}(50-t)^{\small{-\frac{5}{2}}}A=(50-t)^{\small{-\frac{3}{2}}}$ 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: $\frac{d}{dt}$$(50-t)^{\small{-\frac{3}{2}}}A$$=(50-t)^{\small{-\frac{3}{2}}}\frac{dA}{dt}+\frac{3}{2}(50-t)^{\small{-\frac{5}{2}}}A$ Hence we may now write the ODE as: $\frac{d}{dt}$$(50-t)^{\small{-\frac{3}{2}}}A$$=(50-t)^{\small{-\frac{3}{2}}}$
 February 29th, 2012, 06:26 PM #5 Member   Joined: Sep 2011 Posts: 97 Thanks: 0 Re: expression in mixing problem $\frac{d}{dt}$$(50-t)^{\small{-\frac{3}{2}}}A$$=(50-t)^{\small{-\frac{3}{2}}}\frac{dA}{dt}+\frac{3}{2}(50-t)^{\small{-\frac{5}{2}}}A$ 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 $\frac{3}{2}(50-t)^{\small{-\frac{5}{2}}}$ 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: $\frac{d}{dx}$$f(x)\cdot g(x)$$=f(x)\cdot\frac{d}{dx}$$g(x)$$+\frac{d}{dx}\ (f(x)\)\cdot g(x)$ or using prime notation: $$$f(x)\cdot g(x)$$'=f(x)\cdot g'(x)+f(x)\cdot g#39;(x)$ 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 $f(x)= (50 - t)$ 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 $f(t)=(50-t)^{\small{-\frac{3}{2}}}$ $g(t)=A(t)$

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