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Homework Help: Which difference equation better describes the problem ?

  1. May 1, 2012 #1
    http://dl.dropbox.com/u/33103477/Nicotine.png [Broken]

    [tex] N_{t+1} = N_t - \frac{1.02}{100}N_t + 0.02 [/tex]


    [tex] N_{t+1} = N_t - \frac{1}{100}N_t + 0.02 [/tex]

    The first equation considers counts the amount the body absorbs everyday to include the amount ingested that day, the other doesn't.

    The answers are close but different.

    Which one do you think is correct.

    Also, what does "closed form of the solution" mean ?
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. May 1, 2012 #2


    Staff: Mentor

    The two equations can be rewritten as:

    Nt+1 = -.02Nt + .02
    Nt+1 = .99Nt + .02

    Which one do you think better describes your situtation?
    Last edited by a moderator: May 6, 2017
  4. May 1, 2012 #3
    You've made a mistake there:

    [tex] 1 - \frac{1.02}{100} ≠ -0.02 [/tex]

    I think you though 1 was 1/100.

    In other words the equations are:

    [tex] t_{n+1} = 0.9898t_{n} + 0.02 [/tex]


    [tex] t_{n+1} = 0.99t_{n} + 0.02 [/tex]
  5. May 1, 2012 #4


    Staff: Mentor

    Right - I subtracted 1.02 from 100 and got -.02. I occasionally make a mistake in my checkbook, too!

    So with your correction, which equation looks like the right one?
  6. May 1, 2012 #5
    They both seem good to me, as both "make sense", the steady state for the first is 1.96 and second is 2. The final answers work out close enough to each other.

    The reason I'm putting it up here is to consider other people's interpretation. Would you consider the nicotine added today to be removed by the body today itself ?

    IMO it would really depend on when you take a piss really. If you take a weee in the morning then the nicotine leaves then, take a weee at night nicotine leaves then....eh ?
    Last edited: May 1, 2012
  7. May 1, 2012 #6

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    There are several possibilities: (1) the input 0.02 mg occurs at the start of a day; (2) it occurs at the end of a day; (3) it is spread out evenly throughout the day. Case (3) is trickier, so let's leave it (it makes a good modelling exercise, however). Assume that N_t = amount of nicotine (in mg) right at the start of day t (before any smoking, etc.)

    Case (1): [itex] N_{t+1} = 0.99 (N_t + 0.02). [/itex]

    Case (2): [itex] N_{t+1} = 0.99 N_t + 0.02. [/itex]

    By "closed-form" solution, the question means for you to find a formula N_t = f(t), and it wants the function f.

    Last edited by a moderator: May 6, 2017
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