Results 1 to 3 of 3

Math Help - calc extra credit wokr, i need it!!!

  1. #1
    Newbie
    Joined
    Dec 2007
    Posts
    4

    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!!!!!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    GAMMA Mathematics
    colby2152's Avatar
    Joined
    Nov 2007
    From
    Alexandria, VA
    Posts
    1,172
    Awards
    1
    \frac{dP}{dt}=k\sqrt{t}

    P(0) = 500

    P(1) = 600

    P(7) = ???

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

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

    dP=k\sqrt{t}dt

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

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

    500=C

    100=\frac{2k}{3}

    k = 150

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

    P(7)=2352
    Last edited by colby2152; December 18th 2007 at 04:24 AM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,710
    Thanks
    628
    Hello, Mr.Obson!

    The rate of growth \frac{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 \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: . \frac{dP}{dt} \:=\:k\!\cdot\!\sqrt{t}\quad\Rightarrow\quad dP \:=\:k\!\cdot\!t^{\frac{1}{2}}

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

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

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

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


    When t = 7\!:\;\;P \;=\;100\!\cdot\!7^{\frac{3}{2}} + 500 \;\approx\;2352
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. extra credit
    Posted in the Pre-Calculus Forum
    Replies: 3
    Last Post: March 30th 2009, 10:10 PM
  2. Calc 2 extra credit problem
    Posted in the Calculus Forum
    Replies: 4
    Last Post: March 17th 2009, 04:38 AM
  3. geometry extra credit help
    Posted in the Geometry Forum
    Replies: 4
    Last Post: November 16th 2008, 08:10 PM
  4. extra credit
    Posted in the Calculus Forum
    Replies: 1
    Last Post: March 4th 2008, 06:00 PM
  5. extra credit help
    Posted in the Math Topics Forum
    Replies: 14
    Last Post: December 19th 2007, 03:14 AM

Search Tags


/mathhelpforum @mathhelpforum