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Math Help - Growth/Decay problem

  1. #1
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    Growth/Decay problem

    I have the following question from a past paper which I seemed to be rusty on and just wanted to go through it:

    A virus spreads slowly through a population of animals. Initially no animals are infected, but after 10 days 25% of the population in infected. Let P(t) denote the proportion of the population that is infected. Then

    \frac{dP}{dt} = k(1-P)

    (i) Solve the differential equation to get an expression for P(t).
    (ii) Find the value of the constant k.
    (iii) How long will it take for 75% of the population to become infected?

    My attempt:

    (i)

    \integral \frac{dP}{dt} = \integral k(1-P)

    dP=k(1-P)~dt

    \int \frac{1}{1-P}~dP = \int k~dt

    -ln|1-P| = kt~dt

    -1+P = e^{kt}

    P(t) = e^{kt} +1

    This seems wrong as at P(0), no animals should be infected, but the equation I derived gives a value of 2. What have I done wrong?

    Also, why don't I have a coefficient in front of the e^{kt} term? I have also never seen a constant next to the normal P_0e^{kt} term; is that normal?

    I will wait till we have figured out part (i) before I attempt the next two parts.

    Thanks.
    Last edited by terrorsquid; November 21st 2011 at 01:54 AM.
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  2. #2
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    Re: Growth/Decay problem

    Quote Originally Posted by terrorsquid View Post
    I have the following question from a past paper which I seemed to be rusty on and just wanted to go through it:

    A virus spreads slowly through a population of animals. Initially no animals are infected, but after 10 days 25% of the population in infected. Let P(t) denote the proportion of the population that is infected. Then

    \frac{dP}{dt} = k(1-P)

    (i) Solve the differential equation to get an expression for P(t).
    (ii) Find the value of the constant k.
    (iii) How long will it take for 75% of the population to become infected?

    My attempt:

    (i)

    \integral \frac{dP}{dt} = \integral k(1-P)

    dP=k(1-P)~dt

    \int \frac{1}{1-P}~dP = \int k~dt

    -ln|1-P| = kt~dt

    -1+P = e^{kt}

    P(t) = e^{kt} +1

    This seems wrong as at P(0), no animals should be infected, but the equation I derived gives a value of 2. What have I done wrong?

    Also, why don't I have a coefficient in front of the e^{kt} term? I have also never seen a constant next to the normal P_0e^{kt} term; is that normal?

    I will wait till we have figured out part (i) before I attempt the next two parts.

    Thanks.
    When integrating you have to include the arbitrary constant of integration '+ C'. Get its value from the initial condition.
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  3. #3
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    Re: Growth/Decay problem

    Is this what you mean?

    -ln|1 -P| = kt + C

    -1+P=e^{kt}e^{C}

    let~e^C = P_0

    P(t) = P_0e^{kt}+1

    P(0) = P_0 + 1

    Not sure how to proceed... Initially 0 are infected, so that means P_0 = -1 ?
    Last edited by terrorsquid; November 21st 2011 at 01:15 PM.
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    Re: Growth/Decay problem

    \int \frac{dP}{1-P} = \int k \, dt

    \ln|1-P| = -kt + C

    1 - P = e^{-kt+C} ... let A = e^C

    1 - P = Ae^{-kt}

    P = 1 - Ae^{-kt}

    when t = 0 , P = 0 (which is very unrealistic for a problem like this)

    A = 1

    P = 1 - e^{-kt}

    at t = 10, P = 0.25

    0.25 = 1 - e^{-10k}

    solve for k and finish it.
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    Re: Growth/Decay problem

    Just out of curiosity, how did you spot to switch the sides of kt and ln|1-P|? After considering the situation presented by the question, it seems we are dealing with an exponential growth problem which would suggest I need to find a formula with a +k value. Typically I would try to get it into the form of +k for growth and -k for decay. I can see now with the final answer that it makes sense, but how would you spot it before getting to the end?
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  6. #6
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    Re: Growth/Decay problem

    Quote Originally Posted by terrorsquid View Post
    Just out of curiosity, how did you spot to switch the sides of kt and ln|1-P|? After considering the situation presented by the question, it seems we are dealing with an exponential growth problem which would suggest I need to find a formula with a +k value. Typically I would try to get it into the form of +k for growth and -k for decay. I can see now with the final answer that it makes sense, but how would you spot it before getting to the end?
    \int \frac{1}{1-P} \, dP = \int k \, dt

    \int \frac{-1}{1-P} \, dP = \int -k \, dt

    ... since the derivative of (1-P) is -1
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  7. #7
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    Re: Growth/Decay problem

    So, whenever I have a problem where the term being multiplied by the constant k in the instantaneous growth/decay rate equation: \frac{dP}{dt}=k(term)
    is negative, I look to find a P(t) term with a -k exponent and vice-versa, regardless of whether the question is about growth or decay?

    Continuing:

    (i)

    0.25 = 1-e^{-10k}

    -0.75 = -e^{-10k}

    e^{-10k}=0.75

    -10k = ln(0.75)

    k = 0.02876820725

    (ii)

    0.75 = 1-e^{-0.02876820725t}

    t = \frac{ln(0.25)}{-0.0287682072}

    =48.19~days
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  8. #8
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    Re: Growth/Decay problem

    you're asking me "do I always?" ... understand that I cannot say yes. each problem is different. I just took advantage of being able to multiply both sides by -1 to make the antiderivative easier to find.
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