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Math Help - Calculus Help Please

  1. #1
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    Calculus Help Please

    Initially a population contains 1000 individuals. After t minutes the number N of individuals in the populataion is such that dN/dt = -0.25N.
    a) write down the expression for N in terms of t
    b) Find the rate at which the population is decreasing when it is half its orginal size.

    This question is driving me nuts Any help would be much appreciated thanks

    Reguards, Luke
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Luke007 View Post
    Initially a population contains 1000 individuals. After t minutes the number N of individuals in the populataion is such that dN/dt = -0.25N.
    a) write down the expression for N in terms of t
    b) Find the rate at which the population is decreasing when it is half its orginal size.

    This question is driving me nuts Any help would be much appreciated thanks

    Reguards, Luke
    a)

    \frac {dN}{dt} = -0.25N

    \Rightarrow \frac {dN}{N} = -0.25 dt

    \Rightarrow \ln N = -0.25t + C

    \Rightarrow N = e^{-0.25t + C}

    \Rightarrow N = e^C e^{-0.25t}

    \Rightarrow N = Ae^{-0.25t} ......we could have jumped straight to here, but I wanted to show you how we got here. This formula should be in your book

    when t = 0, N = 1000

    \Rightarrow N(0) = Ae^0 = 1000

    \Rightarrow A = 1000

    \Rightarrow N(t) = 1000e^{-0.25t}


    b)
    \frac {dN}{dt} = -0.25N

    when N is half it's size:

    \frac {dN}{dt} = -0.25 (0.5N)

    \frac {dN}{dt} = -0.125N

    So the population is decreasing at a rate of -0.125
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  3. #3
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    You're a life saver Jhevon, this will help with that question and the other ones in my text book, thank you very much

    Reguards, Luke

    PS: Thanks for the expanding on part a) It'll really help in the long run so I understand how you did it
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Luke007 View Post
    b) Find the rate at which the population is decreasing when it is half its orginal size.
    actually now that i think about part (b), i think i probably would do it a different way. this way i get the answer to be -125, which is a thousand times what we got the answer to be last time, so we can't get away with, "no, professor, that's what i meant to write, it may just be a rounding off error, i deserve full credit!" hopefully this is right.

    Now, for an exponential decay, which is what is happening here, this population seems to be dying out, we can find the half-life, that is, the time taken for half the original amount to remain, using the following formula:

    t_h = \frac { \ln 2}{r}, where t_h is the half-life, r is the rate of decrease. but let's see how this formula came about in the first place.

    We want to find first, the time when the population reaches half its original amount, that is, when N = 500

    we want 500 = 1000e^{-0.25t}

    \Rightarrow \frac {1}{2} = e^{-0.25t}

    \Rightarrow \ln \left( \frac {1}{2} \right) = \ln e^{-0.25t}

    \Rightarrow - \ln 2 = -0.25t

    \Rightarrow t = \frac { \ln 2}{0.25} ...look familiar?


    Now N = 1000e^{-0.25t}

    \Rightarrow \frac {dN}{dt} = -250e^{-0.25t}

    when t = \frac { \ln 2}{0.25}:

    \frac {dN}{dt} = -250e^{-0.25 \cdot \frac { \ln 2}{0.25}}

    \Rightarrow \frac {dN}{dt} = -250 \cdot \frac {1}{2}

    \Rightarrow \frac {dN}{dt} = -125
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