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Math Help - Salt Mixture Problem (First-Order Linear ODE Problem)

  1. #1
    Super Member Aryth's Avatar
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    Salt Mixture Problem (First-Order Linear ODE Problem)

    I just want to verify my work... Hopefully I was right:

    A tank initially contains 100 gallons of brine in which 50 lbs. of salt are dissolved. A brine containing 2 lbs/gal of salt runs into the tank at the rate of 5 gal/min. The mixture is kept uniform by stirring and flows out of the tank at 4 gal/min.

    Initial Information:

    IN: 2 lbs/gal AT: 5 gal/min
    OUT: 4 gal/min

    V(t) = Volume

    y(t) = Concentration

    \frac{dy}{dt} = \text{RATE IN} - \text{RATE OUT}

    a) At what rate (lbs/min) does the salt enter the container at time t?

    \text{RATE IN} = 2\frac{lbs}{gal}*5\frac{gal}{min} = 10\frac{lbs}{min}

    b) What is the volume of brine in the tank at time t?

    V(t) = 100 + t

    c) At what rate (lbs/min) does salt leave the tank at time t?

    \text{RATE OUT} = 4*\left(\frac{y(t)}{100 + t}\right)

    d) Write and solve initial value problem.

    \frac{dy}{dt} = 10 - \left(\frac{4}{100 + t}\right)y

    \frac{dy}{dt} + \left(\frac{4}{100 + t}\right)y = 10

    IF = (100 + t)^4

    [(100 + t)^4y]' = 10(100 + t)^4

    \int [(100 + t)^4y]' = \int 10(100 + t)^4 dt

    (100 + t)^4y = \frac{(100 + t)^5}{2} + C

    y = \frac{100 + t}{2} + C

    y(0) = 50 + C = 0

    C = -50

    y = \frac{100 + t}{2} - 50

    y = \frac{t}{2}

    e) Find the concentration of salt in the tank at 25 minutes after process starts.

    y(25) = \frac{25}{2}

    y = 12.5 lbs.
    Last edited by Aryth; February 29th 2008 at 09:13 PM.
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  2. #2
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    Quote Originally Posted by Aryth View Post
    I just want to verify my work... Hopefully I was right:

    A tank initially contains 100 gallons of brine in which 50 lbs. of salt are dissolved. A brine containing 2 lbs/gal of salt runs into the tank at the rate of 5 gal/min. The mixture is kept uniform by stirring and flows out of the tank at 4 gal/min.

    Initial Information:

    IN: 2 lbs/gal AT: 5 gal/min
    OUT: 4 gal/min

    V(t) = Volume

    y(t) = Concentration

    \frac{dy}{dt} = \text{RATE IN} - \text{RATE OUT}

    a) At what rate (lbs/min) does the salt enter the container at time t?

    \text{RATE IN} = 2\frac{lbs}{gal}*5\frac{gal}{min} = 10\frac{lbs}{min}

    b) What is the volume of brine in the tank at time t?

    V(t) = 100 + t

    c) At what rate (lbs/min) does salt leave the tank at time t?

    \text{RATE OUT} = 4*\left(\frac{y(t)}{100 + t}\right)

    d) Write and solve initial value problem.

    \frac{dy}{dt} = 10 - \left(\frac{4}{100 + t}\right)y

    \frac{dy}{dt} + \left(\frac{4}{100 + t}\right)y = 10

    IF = (100 + t)^4

    [(100 + t)^4y]' = 10(100 + t)^4

    \int [(100 + t)^4y]' = \int 10(100 + t)^4 dt

    (100 + t)^4y = \frac{(100 + t)^5}{2} + C *

    Mr F says: This line should be (100 + t)^4y = {\color{blue}2} (100 + t)^5 + C. But that's the least of your worries ......

    y = \frac{100 + t}{2} + C Mr F says: No. It's y = \frac{100 + t}{2} + \frac{C}{(100 + t)^4}. So I'm afraid that everything that follows is no good.

    y(0) = 50 + C = 0

    C = -50

    y = \frac{100 + t}{2} - 50

    y = \frac{t}{2}

    e) Find the concentration of salt in the tank at 25 minutes after process starts.

    y(25) = \frac{25}{2}

    y = 12.5 lbs.
    Two things:

    1. y is the amount (in lb) of salt, not concentration.

    2. I'm afraid your integration goes South after the line marked *.

    You could have seen that your solution was no good by substituting it back into the original DE ...... It doesn't work.
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