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Math Help - Population Models

  1. #1
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    Population Models

    Okay I have been doing this for a while now and I got t = 3, but clearly it was wrong because it actually equals 1.6 days.

    I set up the model like this

    dp/dt = k(p)^1/2


    This is the problem

    Let t denote time (in days) and let P denote the population of some species at time t. Suppose that the birth rate β (number of births per day per unit of population) is proportional to P^1/2 and that the death rate δ (number of deaths per day per unit of population) equals 0. Suppose the initial population is 4, and after one day the population is 16. When will the population equal 100?
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  2. #2
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    Quote Originally Posted by radioheadfan View Post
    Okay I have been doing this for a while now and I got t = 3, but clearly it was wrong because it actually equals 1.6 days.

    I set up the model like this

    dp/dt = k(p)^1/2


    This is the problem

    Let t denote time (in days) and let P denote the population of some species at time t. Suppose that the birth rate β (number of births per day per unit of population) is proportional to P^1/2 and that the death rate δ (number of deaths per day per unit of population) equals 0. Suppose the initial population is 4, and after one day the population is 16. When will the population equal 100?
    Please show all your working that led to your answer of t = 3. Every step, please.
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  3. #3
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    okay I'm guessing i setup the model correctly
    So, I took the integrals of (1/(p)^(1/2)) dP = k dt

    and I got 2p^(1/2) = kt + c

    p = ((kt/2) + c)^2

    c = Po^1/2 (initial population)

    so i plugged in the initial values to find the constant k value

    16 = ( k/2 + (4)^(1/2))^2 ... Took the square root of both sides

    4 = (k/2 + 2)

    2 = k/2

    4 = k

    then for the new value of P (100) i plugged in the values

    100 = ( 4t/2 + (4)^1/2)^2

    square root

    10 = 2t +2

    8 = 2t

    4 = t

    I actually got 4, but the Professor said it was equal to 1.6 days.
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  4. #4
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    Quote Originally Posted by radioheadfan View Post
    okay I'm guessing i setup the model correctly
    So, I took the integrals of (1/(p)^(1/2)) dP = k dt

    and I got 2p^(1/2) = kt + c

    p = ((kt/2) + c)^2

    c = Po^1/2 (initial population)

    so i plugged in the initial values to find the constant k value

    16 = ( k/2 + (4)^(1/2))^2 ... Took the square root of both sides

    4 = (k/2 + 2)

    2 = k/2

    4 = k

    then for the new value of P (100) i plugged in the values

    100 = ( 4t/2 + (4)^1/2)^2

    square root

    10 = 2t +2

    8 = 2t

    4 = t

    I actually got 4, but the Professor said it was equal to 1.6 days.
    p = ((kt/2) + c)^2

    Substitute t = 0, p = 4: c = 2.

    Therefore p = ((kt/2) + 2)^2.

    Substitute t = 1, p = 16. Get k = 4.

    p = (2t + 2)^2.

    Substitute p = 100: 100 = (2t + 2)^2 => t = 4.

    Looks like you're correct.
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  5. #5
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    are you sure the model is correctly written according to the word problem?
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  6. #6
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    Quote Originally Posted by radioheadfan View Post
    are you sure the model is correctly written according to the word problem?
    Yes, it is correct according to the wording of the question you posted.
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