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Calculation Problem Solving

  1. Jun 14, 2010 #1
    [I have here a question from one of my many revision sheet in preperation for my exams.
    The problem I'm having is with this parcticular question.
    [/SIZE]

    A 10 year research program is being carried out on two penguin rookery populations in Tasmania. Rookery A had 2000 penguins at the start of the study and the population has been decreasing by 5% per year. Rookery B had 3000 penguins initially and this population has been decreasing by 10% per year.

    a) Calculate how long it will take for the two populations to be the same.
    b) Find this actual figure.

    State any assumptions and use algebra, logarithms and other mathematics means (not a graphics calculator) to assist in your solution.

    I assumed it was to include growth and decay, so i started trying some things with that and got stuck very quickly.

    Any help is greatly appreciated.
     
  2. jcsd
  3. Jun 14, 2010 #2

    lanedance

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    you're correct its to do with exponetial decay... show your work & where you're stuck

    if the population at time t is N(t), you should start by setting a a simple first order DE, for the rate of change of the population dN/dt, in terms of N(t)
     
  4. Jun 14, 2010 #3
    I have the same problem working out this, I dont know where to start, Could someone please help.
     
  5. Jun 14, 2010 #4
    The first thing i attempted to do was find the decay factor using

    f(x) = AxB^X

    so for rookery A (2000 penguin, at a decay rate of 5% each year.)

    But i am unsure on how to apply this to the formula.
     
  6. Jun 14, 2010 #5
    Hey, this question looks like a wopa, can someone pleaseshow the working out?
     
  7. Jun 14, 2010 #6

    HallsofIvy

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    X is number of years? Since any number to the 0 power is 1, when X= 0, f(0)=A= 2000.

    Since this is a calculus problem, you can try using the derivative to get the decay rate equal to .05.

    By the way, when you are using "X" as a variable, it is not a good idea to use "x" to represent multiplication. Use parentheses: f(x)= A(B^X).
     
  8. Jun 14, 2010 #7

    lanedance

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    ok with Rookery A, you know
    [tex] N_A(0) = 2000 [/tex]
    [tex] N_A(1) = 0.95 N_A(0) [/tex]

    assume an exponetial decay [itex] N_A(t) = A_0 e^{-ct}[/tex], and use the constraints above to solve for the constants
     
  9. Jun 14, 2010 #8
    Assuming that the decay rate for each population does not change before the populations are equal:

    (a) A = 2000(1 - 0.05)^n

    So A = 2000 * 0.95^n [eqn 1]

    B = 3000 * 0.90^n

    Now pops are the same when A = B

    so 2000 * 0.95^n = 3000 * 0.90^n

    0.95^n / 0.90^n = 3/2

    Taking logs to base 10 of both sides:

    log (0.95^n / 0.90^n) = log (3/2)

    log 0.95^n - log 0.90^n = log (3/2)

    n log 0.95 - n log 0.90 = log (3/2)

    n(log 0.95 - log 0.90) = log (3/2)

    n = log (3/2) / (log 0.95 - log 0.90)

    n = 7.5

    So it will take 7.5 years (to 1 dec pl) for the two populations to be the same



    (b) Subs n = 7.5 into [eqn 1] → A = 2000 * 0.95^7.5

    A = 1361

    So the populations will both be 1361 penguins
     
  10. Jun 14, 2010 #9
    Alrite, cheers jhaddow. Now it makes sense to me. That's all i need to see. thankyou
    thank you thankyou.
     
  11. Jun 14, 2010 #10

    lanedance

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    one step tha might help is as follows
    0.95^n / 0.90^n = 3/2

    then
    (0.95/ 0.90)^n = 3/2

    log of boths sides
    n*log(0.95/ 0.90) = log(3/2)
     
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