1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Rate of increase/differential equation

  1. May 30, 2009 #1
    The Problem:

    For a species of bacterium, the rate of increase of the population is determined by the following information:

    the rate of increase is proportional to the population P(t)

    the rate of increase is proportional to the temperature T(t)

    My question:

    I initially wrote out two seperate equations for this, namely dP/dt = aP(t) and dP/dt = bT(t), then using the initial conditions P(0)=1000 and P(10)=2000 and the information T(t)=40-2t solved each, one giving an exponential and the other a polynomial.

    Should I have instead solved the equation dP/dt = kP(t)T(t)?

    I'm pretty sure the answer is 'yes' but I'd like some confirmation anyway.
     
  2. jcsd
  3. May 30, 2009 #2
    incidentally I ultimately have to find P(t) at t=20, and using the method I just described I get 1000 times the cube root of 16. Is that correct?
     
  4. May 30, 2009 #3
    Yes, solve for dP/dt = kP(t)T(t). Think way back from algebra.
     
  5. May 30, 2009 #4
    Thanks

    I would like confirmation for this also
     
  6. May 30, 2009 #5

    Cyosis

    User Avatar
    Homework Helper

    Giving you confirmation would mean that we would have to solve your entire problem first and then compare it to your answer. Secondly we don't even know to which method your solution belongs, the wrong one or the correct one?

    If you want confirmation show us your work step by step so we can see where and if you made mistakes. We will then help you from thereon.
     
    Last edited: May 30, 2009
  7. May 30, 2009 #6
    Yeah okay.

    P'(t) = k P(t)T(t)

    T=(40-2t)

    P'(t)/P(t) = 40k - 2kt

    P^(-1)dP = (40k -2kt)dt

    lnP = 40kt - kt^2 + C

    P(t) = Ce^(40kt - kt^2)

    P(0) = 1000 => C=1000

    P(t) = 1000e^(40kt - kt^2)

    P(10) = 2000 = 1000e^300k => k = (ln2)/300

    P(t) = 1000e^(((2/15)ln2)t - ((1/300)ln2)T^2)

    P(20) = 1000e^((4/3)ln2)
    P(20) = 1000*(16^(1/3))
     
  8. May 30, 2009 #7

    Cyosis

    User Avatar
    Homework Helper

    Every step looks correct to me. A slightly nicer way to write it would perhaps be 2000 2^(1/3), put that's purely aesthetic.
     
  9. May 30, 2009 #8
    thanks again. I'll probably be posting one or two more of these this weekend
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Rate of increase/differential equation
Loading...