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Help creating a dif. eq. for a real life model?

  1. Feb 6, 2008 #1
    Help creating a dif. eq. for a real life model.help???

    A 30 year old woman accepts an engeineeering position with a starting salary of $30000 per year. Her salary S(t) increases exponentially, with [tex]S(t)=30e^{\frac{t}{20}[/tex] thousand dollars after t years. Meanwhile, 12% of her salary is deposited continuously in a retirement account, which accumulates interest at a continuous annual rate of 6%.
    a) Estimate dA in terms of dt to derive the differential equation satisfied by the amount A(t) in her retirement account after t years.

    Well, i am not any good at differential equations, but i am just trying to give a shot on my own, although i currently am in calculus I, so i am facing some difficulties from time to time absorbing these kind of problems, especially when they are wording ones that require us to derive a differential equation to describe the situation.

    I have tried many things on this problem, first i thought that the differential equation derived for measuring the concentration on a tank- as one pipe brings water in while the other pipes water out of tank- would work but i think it does not quite fit in this situation.

    so any hints on how to begin to derive a diff. eq for describing this situation??

    thnx in advance
     
    Last edited: Feb 6, 2008
  2. jcsd
  3. Feb 7, 2008 #2

    HallsofIvy

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    A(t) increases in two ways:
    1) She deposits 12% of her salary: 0.12 S= 0.12(30 et/20[/itex]= 3.6et/20.
    2) interest, compounded continuously at 6% is added: A(t)e0.06t.
    Those are the two ways in which the account changes and the rate of change is dA/dt.
    Looks to me like dA/dt= e0.06tA(t)+ 3.6 et/20.

     
  4. Feb 7, 2008 #3
    Shouldn't we take into consideration somewhere her starting salary, maybe you incorporated it somewhere but i just cannot se where? If not why would it be ok, even if we do not take it into consideration?
    Also how did you derive A(t)e^0.06t ? In other words, how did you incorporate the 6% interest here?
     
  5. Feb 7, 2008 #4

    Ben Niehoff

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    Geeze, I certainly hope not! That's an obscenely low salary for any 30-year-old, especially an engineer!
     
  6. Feb 7, 2008 #5

    HallsofIvy

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    The initial salary is already taken into account in the S(t)= 30et/20 although we should write it as 30000et/20.

    The 6% is the 0.06 in the exponential. Amount A, compounded continuously, at interest rate r, for t years yields Aert.
     
  7. Feb 7, 2008 #6

    HallsofIvy

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    First college teaching job I got, after I got my Ph.D, paid $12000 a year. And that was more than my father had ever earned in one year in his life.
     
  8. Feb 7, 2008 #7

    Ben Niehoff

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    Yeah, but what decade was that?

    Hell, even my first internship was $18/hr (though I realize I'm a bit lucky there). :P
     
  9. Feb 7, 2008 #8
    Where can i find some things to read on my own about this interest rate, because i do not actually know why amount A, compounded continuously, at interest rate r, for t years yeilds Aert??
     
  10. Feb 7, 2008 #9

    HallsofIvy

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  11. Feb 7, 2008 #10
    Well, thank you HallsofIvy, i really appreciate it. I will come back from time to time with these kind of problems, untill i get used to them.

    Thnx once more!
     
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