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A problem on differential equations

  1. Mar 7, 2016 #1
    1. The problem statement, all variables and given/known data
    A boy is eating a 250 gram burger. When he has eaten a mass m grams, his rate of consumption is given by ## 100-1/900m^2## grams per minute. How long does it take him to finish the meal?



    2. Relevant equations


    3. The attempt at a solution

    ##dm/dt=100-1/900m^2 , dm/dt=(90000-m^2)/900
    ⇒dt/dm=900/(90000-m^2), t=∫900/(300+m)(300-m) dm##
    let
    ##u=(300+m), t=∫900/(u(600-u))du=
    900/(u(600-u))= (A/u +B/u)## on solving, ##A=1.5 ##and ##B=1.5## thus
    ##t=∫(1.5/u+1.5/(600-u))du =1.5lnu+1.5ln(600-u)+k
    =1.5ln(300+m)+1.5ln(300-m)##
    using
    ##t(250)=0## we have
    ## 0=1.5ln(300+250)+1.5ln(300-250)+k## where k=-1.5ln27500##
    thus

    ## t=1.5ln(300+m)+1.5ln(300-m)-1.5ln27500
    at ##m=0##

    i assumed that here the burger would have been eaten and no remnants
    ##t=1.5ln300+1.5ln300-1.5ln27500##
    ##t=1.78##
    the text book answer on this is ##t=3.6 minutes##
     
    Last edited: Mar 7, 2016
  2. jcsd
  3. Mar 7, 2016 #2

    SteamKing

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    Is this a 25 gram burger or a 250 gram burger?

    A 25 gram burger is about 1 bite.

    Also, is the consumption rate ##100 - (\frac{1}{900}m^2)## or ##(100 - \frac{1}{900m^2})##
     
  4. Mar 7, 2016 #3
    consumption rate is ##100-(1/900)m^2##
     
  5. Mar 7, 2016 #4

    SteamKing

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    What about the initial size of the burger - 25 grams or 250 grams?
     
  6. Mar 7, 2016 #5
    the initial burger is 250grams. it is eaten (all of it) until it becomes 0 grams.
     
  7. Mar 7, 2016 #6

    SteamKing

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    After this point, you have some errors in your integration for the second partial fraction term.
     
  8. Mar 8, 2016 #7
    I have seen the error:
    ##t=1.5 ln u - 1.5 ln (600-u) + k ##
    eventually

    ## t = 1.5 ln 300 - 1.5 ln 300 - 1.5 ln 11##
    ## t = -3.6min##
    why do we have a negative? is it taking care of the fact that the burger was initially whole and with every bite it kept on decreasing in mass value until the point
    ## m=0## ?
     
  9. Mar 8, 2016 #8

    Samy_A

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    ##m## is the mass eaten:
     
  10. Mar 8, 2016 #9
    am not getting you ## dt/dm=1.5/(300+m)-1.5/(300-m)##
    and which is equal to
    ##1.5/u+1.5/(300-(u-300)## =
    ## 1.5/u-1.5/(600-u)## =
    ##900/(u(600-u))## =
    ##900/(300+m)(300-m)## =
    ## 900/(90000-m^2)## as indicated in my original post. my question still is why do we have ##-3.6##minutes?
     
  11. Mar 8, 2016 #10

    Samy_A

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    Your calculation looks correct to me (I thought there was a sign error, but I was wrong about that).
    Your interpretation of ##m## is wrong. ##m## is the mass eaten, so at t=0, m=0, and when the boy finishes his hamburger, m=250 grams.
     
  12. Mar 8, 2016 #11
    ok sam let me have a look at my equations again, greetings from Africa
     
  13. Mar 8, 2016 #12

    Samy_A

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    To be totally clear, your formula for ##t## is correct, but the error starts when you compute the integration constant ##k##.
     
  14. Mar 8, 2016 #13
    thanks now clear , we have
    ##k=0##
    then
    ## t= 1.5ln 550-1.5ln50##
    =##1.5ln(550/50)= 1.5ln11 = 3.59 ## minutes
     
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