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Differential equations - population growth models

  1. Feb 6, 2007 #1
    The initial mass of a species of fish in a lake is 7 million tonnes. The mass of fish, if left alone, would increase at a rate proportional to the mass, with a proportionality constant of 2/year. However, commercial fishing removes fish mass at a constant rate of 15 million tonnes per year. Answer the following questions.
    (a) When will all the fish be gone?
    (b) What should the fishing rate be so that the mass of fish remains constant?



    I am confused as to how to start this question. i am unsure whether i am suppose to use dp/dt = ap-bp^2 nor do i know how to plug everything in. In class, i was given that p(t) = ap0/[bp0 + (a-bp0)e^(-a(t-t0))], i just do not know how to use it.


    This is my attempt, I just do not know how to start. but I am guessing for (a) that i would have to set somethign = to zero. I am given a = 2, P(0) = 7000000? and b=15?? But I am confused because if i set p(t)=0, then 0=ap0, and both a and p0 are given.... so i KNOW i am doing somethign wrong but do not know how to fix it.
    I just have some new insight on problem (a), perhaps i do not need to use the complicated model but just the easy P=P0e^a(t-t0) where a = 2-15, and p0 is 7mill and t0 is 0.. this makes a lot of sense to me. but then it is insolvable since i end up gett 0=e^-13t so once again i am back to square 1 and confused.
    and i am still confused as to how to start (b) , any help would be greatly appreciated as i have no idea what to do .
     
    Last edited: Feb 6, 2007
  2. jcsd
  3. Feb 6, 2007 #2
    I'm doing this stuff in my differential equations class right now too, so maybe I could help. You should set P(0) = 7, where 7 is in millions of tons. Then it says the proportionality constant is 2, and that commercial fishing removes fish at a constant rate of 15 (million tons) per year, so b = -15. So, after all that, your equation should look like this:

    dp/dt = 2P - 15

    After that, you need to solve this differential equation for P and set it equal to 0 (since you want to find out when the population of fish is equal to 0). Hope that helps!
     
  4. Feb 7, 2007 #3
    okay so i have done waht you said
    and i am stuck
    also, why is b=-15 but never used anyhwere, i am not really sure how you got dp/dt=2P-15 because isnt the logistic model dp/dt=ap-bp^2

    so doing it your way, i am getting this.
    dp/dt = 2p-15
    1/2p-15 dp/dt = 1
    integrating both sides with definite integral (p to p=7, and t to t=0)
    i got
    [1/2 ln |2p-15|] (p to 7) = t (t to 0)
    so
    1/2 ln |2p-15| - 1/2 ln |2(7)-15| = t - 0
    1/2 ln |2p-15| - 1/2 ln |-1| = t (okay this is the part where i am stuck, because it is absolute value, will that turn into a +1, i assumed this is what happens. and continued along)
    1/2 ln |2p-15| - 0 = t
    1/2 ln |2p-15| = t
    so now i solve for p
    (2p-15)^1/2 = e^t
    2p-15 = e^2t
    2p = 15 + e^2t
    p = (15 + e^2t)/2
    so i set p = 0 for (a)
    and i get
    15 + e ^2t = 0
    e^2t = -15
    and now... i am stuck. e^2t would never equal - 15... i know i have done something wrong.
     
  5. Feb 7, 2007 #4
    You have 1/2ln|2p-15| = t, put p=0 and remember you are taking the absolute value.
     
  6. Feb 7, 2007 #5
    Sorry I even mentioned setting b = -15. Forget I even said that, and forget about "b". As far as the logistic model goes, there is not one set model for every single problem. The model for this particular problem happens to be dp/dt = 2p(t) - 15. This says that the population changes at a rate proportional to the mass (dp/dt = p(t)) with a proportionality constant 2 (dp/dt = 2p(t)) and that 15 million tons are taken out by fishing (dp/dt = 2p(t) - 15). Understand how I got that?
     
  7. Feb 8, 2007 #6
    yes i understand how you got that now. thanks for the great explanation, i just always assumed that we would not have to come up with our own differential equation as the prof kinda leaned towards that, but apparntly not =/

    and following what jing said,
    i get
    1/2ln|2p-15| = t
    sub p=0
    1/2 ln |2(0) -15| = t
    1/2 ln |-15| = t
    t = 1/2 ln 15

    as for #b, it says we need to find what the fishing rate is, so would this be the time to introduce b where b is unknown? this is very confusing, i am unsure of when to use both a and b. can someone please clarify b for me, i dont know where to begin. thank you
     
  8. Feb 8, 2007 #7
    Part (a) looks like it's correct. For part b, the question is asking what the fishing rate needs to be to make the mass of fish constant, or dp/dt = 0. I'm not sure if this means that the fishing rate should be 2p (i.e. dp/dt = 2p - 2p = 0) or not. That seems almost too easy.
     
  9. Feb 8, 2007 #8
    yeah i actually thought of that too but i didnt think it would be that easy
    anyone else have any thoughts on this one???
     
  10. Feb 8, 2007 #9
    (b) What should the fishing rate be so that the mass of fish remains constant?

    i read the question again and thought of it, and i really belive that the answer is 2p, i feel like it's telling me to set dp/dt = 0 and solve, which means it equals 2p. any thoughts????
     
  11. Feb 10, 2007 #10
    What is the value of p? Think about initial value
     
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