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Math Equation

  1. Jan 1, 2005 #1
    Alright, my teacher the sadist gave us an assignment over the holidays, and so far, everyone in my class has gotten all but this.
    The equation is representative of the population growth of fish in a fish pond.

    p= m / 1+ae to the -kt


    p is the intial population, m the carrying capacity of the pond, t is time, k is a constant of 0.9, and nobody knows what a is. The question is basically, is a a constant, or what? I'm not worried about solving everthing, just whether a is a constant, or whatever, I think a is the number of additional fish every year, but I'm not sure. If it's any help, I've found the carrying capacity to be 53869 fish, and the intial population is 1000.
     
  2. jcsd
  3. Jan 1, 2005 #2
    so its p=[(m)/(1+ae)]^-kt ? 1000 = [(53869)/(1+ae)]^-.9t

    Can you copy the enitre problem?

    what is e supposed to represent?
     
  4. Jan 1, 2005 #3

    dextercioby

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    I'm sorry,i didn't get your formula right.I would have liked u had written it using 'latex'.

    If it is something like that:
    [tex] p=\frac{m}{1+a\exp(-kt)} [/tex]
    ,then 'a' cannot be number of fish,because it needs to be adimensional.'k' has the dimension of time^{-1} and 'p' and 'm' have the meaning of number of fish,therefore adimensional.

    Daniel.

    PS.Please state the question in initial form.And try to put the equation,if not in 'tex',then in words.E.g.the eq.i wrote would be:"p" is:fraction line,at the numerator "m",at the denominator 1 plus exponential of minus 'kt'.Everybody would understand that.On normal basis...
     
    Last edited: Jan 1, 2005
  5. Jan 2, 2005 #4
    Apologies for not being clear enough yesterday, and not getting back until now. e is a constant on our calculators, stated as 2.718281828, and the problem is in the attachment,or here if I did it right.

    The question:
    (a) Given you stock the pond with 1000 fish initially, determine the value of a.
    (b) How many years will it take for your pond to reach its carrying capacity?

    The problem being, if a is a constant, what do we look at as the fresult of the population?
     

    Attached Files:

    Last edited: Jan 2, 2005
  6. Jan 2, 2005 #5

    HallsofIvy

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    Would you mind telling us HOW, given only the information you gave (and the url you give do3es not have ANY problem, just a formula), you found that m= 53869? Assuming that you are given p= m / (1+ae-kt), there were 1000 fish "initially" and, if we assume that t is the time, in years, since the pond was initially stocked (you should have told us what "t" meant- just saying "t is time" isn't enough), then "initially", t= 0 p(0)= m/(1+a) = 53869 /(1+a)= 1000. Can you solve that for a?

    As t gets larger and larger, e-kt approaches 0 so, yes, as t goes to infinity, p approaches m in the limit. Strictly speaking, p is never equal to m but, since p has to be an integer (it is the number of fish in the pond), for what value of t will p= m- 0.5= 53868.5?
    That is, solve 53869/(1+ ae-.9t)= 53868.5 for t, using the value of a you got above.
     
  7. Jan 2, 2005 #6
    t is time in years, sorry, and I solved a for the first part to be 52.869, and from what your saying, p doesn't stay the initial population, it changes as soon as t is greater than 0, therefore, I solve for t, with p equal to 52868.5 aaaand I'm just restating everything you just said, but in some way, that means I understand it. As for finding m, that was a whole other section of the assignment itself, involving estimation, trig, and a very skewed diagram. Regardless, thanks for the help.
     
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