Results 1 to 2 of 2

Math Help - Integration (population growth)

  1. #1
    Member
    Joined
    Jan 2009
    Posts
    100

    Integration (population growth)

    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 (t= [0,10]).

    dP/dt= k*sqrt(t)

    The initial size of the population is 500. After 1 day the population has grown to 600. Estimate the population after 7 days.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member vincisonfire's Avatar
    Joined
    Oct 2008
    From
    Sainte-Flavie
    Posts
    469
    Thanks
    2
    Awards
    1
    *\frac{dP}{dt} = k\sqrt{t} *
     \int dP = \int k\sqrt{t}dt
     P(t) = \frac{2k}{3}t^{\frac{3}{2}} + C
     P(0) = 500 = C
     P(1) = 600 = \frac{2k}{3} + 500 \implies k = 150
     P(t) = 100t^{\frac{3}{2}} + 500
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Population growth DE
    Posted in the Differential Equations Forum
    Replies: 5
    Last Post: December 5th 2009, 12:29 PM
  2. Population Growth
    Posted in the Pre-Calculus Forum
    Replies: 4
    Last Post: May 19th 2009, 03:01 PM
  3. Population growth with non-constant growth factor
    Posted in the Differential Equations Forum
    Replies: 0
    Last Post: February 3rd 2009, 05:46 PM
  4. Integration (population growth)
    Posted in the Calculus Forum
    Replies: 1
    Last Post: January 25th 2009, 02:32 PM
  5. Population Growth
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: December 1st 2008, 12:47 AM

Search Tags


/mathhelpforum @mathhelpforum