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Math Help - Growth problem? Any help would be greatly appreciated!

  1. #1
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    Growth problem? Any help would be greatly appreciated!

    thanks!
    Last edited by camjenson; November 11th 2012 at 10:44 PM.
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  2. #2
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    Re: Growth problem? Any help would be greatly appreciated!

    Hello, camjenson!

    A colony of people grows at a rate directly proportional to the size of the population.
    The colony triples every two hours.

    1. (a) Write differential equation that describes the colony's growth.
    . . (b) Find the general solution.

    (a)\;\frac{dP}{dt} \:=\:k\!\cdot\! P


    (b) We have: . \frac{dP}{P} \:=\:k\,dt

    Integrate: . \ln P \:=\:kt + c \quad\Rightarrow\quad P \:=\:e^{kt+c} \:=\:e^{kt}\!\cdot\! e^c} \:=\:e^{kt}\cdot C

    Hence: . P \:=\:Ce^{kt}

    When t = 0,\;P = P_o, the initial population.

    So we have: . P_o \:=\:Ce^{0} \quad\Rightarrow\quad C \,=\,P_o

    Therefore, the equation is: . P \;=\;P_oe^{kt}




    2. What's the value of the constant of proportionality?

    The population triples every two hours.
    When t = 2,\;P = 3P_o

    We have: . P_oe^{2k} \:=\:3P_o \quad\Rightarrow\quad e^{2k} \,=\,3 \quad\Rightarrow\quad 2k \,=\,\ln3

    . . . . . . . . k \:=\:\tfrac{1}{2}\ln 3 \quad\Rightarrow\quad k \:=\:\ln\left(3^{\frac{1}{2}}\right) \quad\Rightarrow\quad \boxed{k \:=\:\ln\left(\sqrt{3}\right)}




    3. At t = 9 hours, the population is 800 million, what is the initial population?

    The equation is: . P \;=\;P_o\,e^{\frac{1}{2}\ln(3)\cdot t} \;=\;P_o\left(e^{\ln 3}\right)^{\frac{1}{2}t}

    And we have: . P \;=\;P_o\!\cdot\!3^{\frac{1}{2}t}


    When t = 9,\:P = 800,\!000,\!000.

    . . P_o\!\cdot\!3^{4.5} \:=\:800,\!000,\!000

    . . P_o \;=\;\frac{800,\!000,\!000}{3^{4.5}} \;\approx\; 5,\!702,\!225
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