Results 1 to 5 of 5

Thread: Help with a long applied DE problem.

  1. #1
    Super Member Showcase_22's Avatar
    Joined
    Sep 2006
    From
    The raggedy edge.
    Posts
    782

    Help with a long applied DE problem.

    Consider two tanks, A and B, each holding 200 litres of water. A pipe pumps water from tank A to tank B at a rate of 5l/min. At the same time another pipe pumps liquid from tank B to tank A at the same rate. At time t=0, $\displaystyle x_0$kg of a chemical X is dissolved into tank A, and tank B has $\displaystyle y_0$kg of the same chemical X dissolved into it.
    i). Write down the system of differential equations satisfied by x(t) and y(t), the quantity of the chemical X in tanks A and B respectively.
    I put this:

    $\displaystyle x(t)=\frac{-5x_0}{200}+\frac{5y_0}{200}=\frac{1}{40}(-x_0+y_0)$
    $\displaystyle y(t)=\frac{5x_0}{200}-\frac{5y_0}{200}=\frac{1}{40}(x_0-y_0)$

    ii). Find the eigenvalues and the eigenvectors of the resulting matrix form.
    $\displaystyle \frac{1}{40}\begin{pmatrix}
    {-1}&{1}\\
    {1}&{-1}
    \end{pmatrix}\begin{pmatrix}
    {x_0}\\
    {y_0}
    \end{pmatrix}=\begin{pmatrix}
    {x(t)}\\
    {y(t)}
    \end{pmatrix}$

    $\displaystyle \frac{1}{40}\begin{vmatrix}
    {-1-\lambda}&{1}\\
    {1}&{-1-\lambda}
    \end{vmatrix}=\frac{1}{40}((1+\lambda)^2-1)=0$

    $\displaystyle (1+\lambda^2)-1=0$
    $\displaystyle \lambda^2+2\lambda=0 \Rightarrow \lambda(\lambda+2)=0$

    The eigenvalues are 0 and -2.

    $\displaystyle \lambda=0$, Eigenvector:$\displaystyle \begin{pmatrix}
    {-1}\\
    {1}
    \end{pmatrix}$

    $\displaystyle \lambda=-2$, Eigenvector: $\displaystyle \begin{pmatrix}
    {1}\\
    {1}
    \end{pmatrix}$

    iii). Show that the amount of the chemical X in either tank approaches $\displaystyle \frac{1}{2}(x_0+y_0)$ as t approaches infinity.
    $\displaystyle X(t)=A\begin{pmatrix}
    {-1}\\
    {1}
    \end{pmatrix}+Be^{-2t}\begin{pmatrix}
    {1}\\
    {1}
    \end{pmatrix}$

    At t=0:

    $\displaystyle x(0)=x_0$
    $\displaystyle y(0)=y_0$

    $\displaystyle x_0=B-A$
    $\displaystyle y_0=A+B$

    Working this out gives:

    $\displaystyle B=\frac{1}{2}(x_0+y_0)$
    $\displaystyle A=\frac{1}{2}(y_0-x_0)$

    $\displaystyle X(t)=\frac{1}{2}(y_0-x_0)\begin{pmatrix}
    {-1}\\
    {1}
    \end{pmatrix}+\frac{1}{2}(x_0+y_0)e^{-2t}\begin{pmatrix}
    {1}\\
    {1}
    \end{pmatrix}$

    as $\displaystyle t \rightarrow \infty$:

    $\displaystyle X(t) \rightarrow \begin{pmatrix}
    {\frac{1}{2}(x_0+y_0)}\\
    {\frac{1}{2}(y_0-x_0}
    \end{pmatrix}$

    I get really far but this doesn't work out. It suggests that my inital formulae are wrong but I can't see where my mistake is. If I reversed my A and B my formulae would work, but I can't see how I can do this.

    Help would be appreciated greatly!
    Last edited by Showcase_22; Nov 30th 2008 at 12:30 PM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Moo
    Moo is offline
    A Cute Angle Moo's Avatar
    Joined
    Mar 2008
    From
    P(I'm here)=1/3, P(I'm there)=t+1/3
    Posts
    5,618
    Thanks
    6
    Hello,

    Just a remark (or a correction) :
    I put this:

    $\displaystyle x(t)=\frac{-5x_0}{200}+\frac{5y_0}{200}=\frac{1}{40}(-x_0+y_0)$
    $\displaystyle y(t)=\frac{5x_0}{200}-\frac{5y_0}{200}=\frac{1}{40}(x_0-y_0)$
    How are these differential equations ?
    There is not even t in these expressions


    This is how I see the problem.
    Let's care about tank A first.
    There are initially $\displaystyle x_0$ kg of X. So the ratio chemical/litre is $\displaystyle \frac{x_0}{200}$
    Hence if you remove a certain amount of litres, $\displaystyle \alpha$, there will be $\displaystyle \alpha \cdot \frac{x_0}{200}$ that will be removed.

    Now, assuming that t is in minutes... We know that 5 litres are removed within 1 minutes.
    So during t minutes, there are $\displaystyle 5t$ litres that are removed.

    Can you see that if we consider the amount of chemical X removed during t, $\displaystyle \alpha=5t$ ?

    So for tank A, there are $\displaystyle \frac{x_0}{40} t$ kg of X that will be removed and go into tank B.



    Similarly, there are $\displaystyle \frac{y_0}{40}t$ kg of X that will be removed from tank B and that will go into tank A.



    So your equations are :
    (don't forget that there is initially a certain amount of chemical !)

    $\displaystyle x(t)=x_0-\frac{x_0}{40}t+\frac{y_0}{40} t=x_0-t \left(\frac{y_0}{40}-\frac{x_0}{40}\right)$

    and $\displaystyle y(t)=y_0-\frac{y_0}{40}t+\frac{x_0}{40} t=y_0-t \left(\frac{x_0}{40}-\frac{y_0}{40}\right)$


    do tell me if it looks incorrect to you
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member Showcase_22's Avatar
    Joined
    Sep 2006
    From
    The raggedy edge.
    Posts
    782
    Okay, I understand where that came from. It does make more sense than what I put, I forgot to incude a time factor.

    I might be doing something wrong, but with those equations in matrix form I get:

    $\displaystyle \begin{pmatrix}
    {1+\frac{t}{40}}&{\frac{-t}{40}}\\
    {\frac{-t}{40}}&{1+\frac{t}{40}}
    \end{pmatrix}\begin{pmatrix}
    {x_0}\\
    {y_0}
    \end{pmatrix}=\begin{pmatrix}
    {x(t)}\\
    {y(t)}
    \end{pmatrix}$

    When I find the determinant I get:

    $\displaystyle \lambda^2-2(1+\frac{t}{40}) \lambda+(1+\frac{t}{40})^2-\frac{t^2}{40^2}=0$

    This doesn't look right :s

    I was thinking that I have to remove t so I just have constants in the 2x2 matrix. However, I don't kno how to do this.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Rhymes with Orange Chris L T521's Avatar
    Joined
    May 2008
    From
    Chicago, IL
    Posts
    2,844
    Thanks
    5
    Quote Originally Posted by Moo View Post
    So your equations are :
    (don't forget that there is initially a certain amount of chemical !)

    $\displaystyle x(t)=x_0-\frac{x_0}{40}t+\frac{y_0}{40} t=x_0-t \left(\frac{y_0}{40}-\frac{x_0}{40}\right)$

    and $\displaystyle y(t)=y_0-\frac{y_0}{40}t+\frac{x_0}{40} t=y_0-t \left(\frac{x_0}{40}-\frac{y_0}{40}\right)$


    do tell me if it looks incorrect to you
    Looks perfect to me...however, there is a slight problem: they're not differential equations!!!!

    Differentiating the two equations will give you the system you're looking for: $\displaystyle \left\{\begin{aligned}
    \dot{x}(t)&=\tfrac{1}{40}\left(x_0-y_0\right)\\
    \dot{y}(t)&=\tfrac{1}{40}\left(y_0-x_0\right)
    \end{aligned}\right.
    $

    ...which happens to be the system [although there's a little something wrong with his system notation] that Showcase_22 used!!
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member Showcase_22's Avatar
    Joined
    Sep 2006
    From
    The raggedy edge.
    Posts
    782
    I think you wrote those the wrong way round (did you? or am I missing something?? =S)

    So if my initial DEs are correct then the mistake must be later on.

    I know i'm close because I get $\displaystyle \frac{1}{2}(x_0+y_0)$ appearing towards the end. I just can't see where the hole in my logic lies.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. help with derivatives (applied problem)
    Posted in the Calculus Forum
    Replies: 7
    Last Post: Mar 15th 2011, 10:20 AM
  2. Applied Trigonometry Problem
    Posted in the Trigonometry Forum
    Replies: 4
    Last Post: Jan 5th 2010, 11:28 AM
  3. Applied max and min problem
    Posted in the Calculus Forum
    Replies: 1
    Last Post: Nov 9th 2009, 10:53 PM
  4. Applied Optimization Problem
    Posted in the Calculus Forum
    Replies: 2
    Last Post: Dec 9th 2008, 09:34 PM
  5. Applied problem
    Posted in the Algebra Forum
    Replies: 1
    Last Post: Sep 1st 2008, 05:44 PM

Search Tags


/mathhelpforum @mathhelpforum