Results 1 to 7 of 7

Math Help - Dilution Problem (Probably Separable)

  1. #1
    Member
    Joined
    Jun 2010
    Posts
    205
    Thanks
    1

    Dilution Problem (Probably Separable)

    So I'm doing this problem 3 on page 125 of Tenenbaum's ODE and I'm not getting the same answer, not sure why.

    It says, suppose you have a tank of 100 gallons of water with 30 lbs of salt, evenly diluted, and 3 gallons are drained per minute while four gallons of fresh water are fed in each minute while constant stirring ensures homogeneity of the fluid's salt content. What is the amount of salt contained after 10 minutes?

    I figure x, t are total quantity of salt and time respectively so \Delta x = -3\Delta t \cdot \frac{x}{100+t} since -3\Delta t will represent the number of gallons of fluid leaving in \Delta t minutes, and \frac{x}{100+t} the quantity of salt per gallon. If that were so, though, when I separate variables and integrate, I wouldn't have any logarithmic functions in my equation, whereas the answer given in the book has an exponential.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Mar 2010
    From
    Florida
    Posts
    3,093
    Thanks
    5
    Quote Originally Posted by ragnar View Post
    So I'm doing this problem 3 on page 125 of Tenenbaum's ODE and I'm not getting the same answer, not sure why.

    It says, suppose you have a tank of 100 gallons of water with 30 lbs of salt, evenly diluted, and 3 gallons are drained per minute while four gallons of fresh water are fed in each minute while constant stirring ensures homogeneity of the fluid's salt content. What is the amount of salt contained after 10 minutes?

    I figure x, t are total quantity of salt and time respectively so \Delta x = -3\Delta t \cdot \frac{x}{100+t} since -3\Delta t will represent the number of gallons of fluid leaving in \Delta t minutes, and \frac{x}{100+t} the quantity of salt per gallon. If that were so, though, when I separate variables and integrate, I wouldn't have any logarithmic functions in my equation, whereas the answer given in the book has an exponential.
    \displaystyle\frac{dA}{dt}=R_{\text{in}}-R_{\text{out}}

    A(0)=30

    R_{\text{in}}=4\text{gals/min of water}\cdot 0\text{lb/gal of salt}=0

    \displaystyle R_{\text{out}}=\frac{A(T)}{100+t}\text{lb/gal}\cdot 3\text{gal/min}

    \displaystyle\frac{dA}{dt}=-\frac{3A}{100+t}\text{lb/gal}

    Is this your equation?

    \displaystyle\frac{dA}{3A}=-\frac{dt}{100+t}

    \displaystyle\int\frac{dA}{3A}=-\int\frac{dt}{100+t}

    \displaystyle\frac{\ln |A|}{3}=-\ln |100+t|+C\Rightarrow\exp{(\ln (A))}=\exp{(-3\ln (100+t))+C_2)}

    \displaystyle A(t)=C_3\exp{(\ln(100+t)^{-3})}=\frac{C_3}{(100+t)^{3}}
    Last edited by dwsmith; January 19th 2011 at 03:05 PM. Reason: changed 1/3 to 3
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Jun 2010
    Posts
    205
    Thanks
    1
    Well gosh-golly god-damnit that's what I got too! (Except at the last step I get an exponent of -3 rather than \frac{-1}{3} since at the previous step we had -3\ln(...).)

    But the book says my function should be such that A(10) = 30e^{-.286}!

    Also, with our function, A(0) = 30 = \frac{C}{(100+t)^{3}} \Rightarrow C = 3/100000 \Rightarrow
    A(T) = \frac{3}{100000(100+t)^{3}} \Rightarrow A(10) = something freaking tiny, which doesn't seem right (even if we used a cube-root it'd be tiny). So it makes me suspect our solution is wrong.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Joined
    Mar 2010
    From
    Florida
    Posts
    3,093
    Thanks
    5
    Quote Originally Posted by ragnar View Post
    Well gosh-golly god-damnit that's what I got too! (Except at the last step I get an exponent of -3 rather than \frac{-1}{3} since at the previous step we had -3\ln(...).)

    But the book says my function should be such that A(10) = 30e^{-.286}!

    Also, with our function, A(0) = 30 = \frac{C}{(100+t)^{3}} \Rightarrow C = 3/100000 \Rightarrow
    A(T) = \frac{3}{100000(100+t)^{3}} \Rightarrow A(10) = something freaking tiny, which doesn't seem right (even if we used a cube-root it'd be tiny). So it makes me suspect our solution is wrong.
    It shouldn't be 1/3. That was an error on my part.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor
    Joined
    Mar 2010
    From
    Florida
    Posts
    3,093
    Thanks
    5
    Quote Originally Posted by ragnar View Post
    Well gosh-golly god-damnit that's what I got too! (Except at the last step I get an exponent of -3 rather than \frac{-1}{3} since at the previous step we had -3\ln(...).)

    But the book says my function should be such that A(10) = 30e^{-.286}!

    Also, with our function, A(0) = 30 = \frac{C}{(100+t)^{3}} \Rightarrow C = 3/100000 \Rightarrow
    A(T) = \frac{3}{100000(100+t)^{3}} \Rightarrow A(10) = something freaking tiny, which doesn't seem right (even if we used a cube-root it'd be tiny). So it makes me suspect our solution is wrong.
    \displaystyle A(0)=\frac{C_3}{(100)^3}=30\Rightarrow C_3=30000000

    \displaystyle A(t)=\frac{30000000}{(100+t)^3}
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Member
    Joined
    Jun 2010
    Posts
    205
    Thanks
    1
    AHHHH, I'm just retarded. That mystery's solved.

    So do you suppose the book is wrong in having an exponential function as part of their solution? Or is there something I'm missing about how our function is related to e?
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor
    Joined
    Mar 2010
    From
    Florida
    Posts
    3,093
    Thanks
    5
    Quote Originally Posted by ragnar View Post
    AHHHH, I'm just retarded. That mystery's solved.

    So do you suppose the book is wrong in having an exponential function as part of their solution? Or is there something I'm missing about how our function is related to e?
    A(10) = 22.5394

    30 exp(-2.86) = 22.5379
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. separable extension and separable polynomial
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: August 30th 2009, 08:22 PM
  2. Dilution Help Please! (M1V1=M2V2) *CHEMISTRY*
    Posted in the Math Topics Forum
    Replies: 4
    Last Post: April 5th 2009, 03:17 PM
  3. Dilution Dilemma
    Posted in the Algebra Forum
    Replies: 4
    Last Post: March 22nd 2009, 12:21 AM
  4. Please help ODE ( separable) one problem'''
    Posted in the Calculus Forum
    Replies: 1
    Last Post: April 24th 2008, 01:55 AM
  5. dilution formula
    Posted in the Math Topics Forum
    Replies: 2
    Last Post: February 12th 2007, 07:17 AM

Search Tags


/mathhelpforum @mathhelpforum