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Thread: calc extra credit wokr, i need it!!!

  1. #1
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    calc extra credit wokr, i need it!!!

    i'm having trouble with this last problem:

    The rate of growth dP/dt of a population of bacteria is proportional to the square root of t, where P is the population size and t is the time in days. (0<=t<=10).

    That is:
    dP/dt = ksqrt(t) sqrt(t) = square root of t, dunno how to type that out with the keyboard.

    The initial size of the population is 500. After 1 day the population has grown to 600. Estimate the population after 7 days.

    any help would be greatly appreciated!!!!!
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  2. #2
    GAMMA Mathematics
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    $\displaystyle \frac{dP}{dt}=k\sqrt{t}$

    $\displaystyle P(0) = 500$

    $\displaystyle P(1) = 600$

    $\displaystyle P(7) = ???$

    There are a variety of ways to solve this problem...

    $\displaystyle \frac{dP}{dt}=k\sqrt{t}$

    $\displaystyle dP=k\sqrt{t}dt$

    Integrate both sides..
    $\displaystyle P(t)=\frac{2k*t^{3/2}}{3}+C$

    $\displaystyle 600=\frac{2k}{3}+C$

    $\displaystyle 500=C$

    $\displaystyle 100=\frac{2k}{3}$

    $\displaystyle k = 150$

    $\displaystyle P(7)=\frac{150*2*(7)^{3/2}}{3}$

    $\displaystyle P(7)=2352$
    Last edited by colby2152; Dec 18th 2007 at 04:24 AM.
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  3. #3
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    Hello, Mr.Obson!

    The rate of growth $\displaystyle \frac{dP}{dt}$ of a population of bacteria
    is proportional to the square root of $\displaystyle t$, where $\displaystyle P$ is the population size
    and $\displaystyle t$ is the time in days. .$\displaystyle (0 \leq t \leq 10)$

    The initial size of the population is 500.
    After 1 day, the population has grown to 600.
    Estimate the population after 7 days.

    We are told: .$\displaystyle \frac{dP}{dt} \:=\:k\!\cdot\!\sqrt{t}\quad\Rightarrow\quad dP \:=\:k\!\cdot\!t^{\frac{1}{2}}$

    Integrate: .$\displaystyle P \;=\;\frac{2}{3}k\!\cdot\!t^{\frac{3}{2}} + C$ .[1]

    When $\displaystyle t=0,\:P = 500\!:\;\;500 \:=\:\frac{2}{3}k\!\cdot\!0^{\frac{3}{2}} + C\quad\Rightarrow\quad C \:=\:500 $

    When $\displaystyle t = 1,\:P = 600\!:\;\;600 \:=\:\frac{2}{3}k\!\cdot\!1^{\frac{3}{2}} + 500\quad\Rightarrow\quad k = 150$

    Substitute into [1]: .$\displaystyle P \:=\:\frac{2}{3}(150)t^{\frac{3}{2}} + 500\quad\Rightarrow\quad P \:=\:100t^{\frac{3}{2}} + 500$


    When $\displaystyle t = 7\!:\;\;P \;=\;100\!\cdot\!7^{\frac{3}{2}} + 500 \;\approx\;2352$
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