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Thread: Growth/ Decay of bacteria popoulation

  1. #1

    Smile Growth/ Decay of bacteria popoulation

    Hi everyone,

    I got a question on a math quest book.

    The questions said in a lake, the population of bacteria will grow at a rate that is proportional to their current population and that in the absence of any outside factors the population will double every 10 days. It is estimates that, on average, (it is an outside factor), on any given day 8% of them will die. The current population is 100.

    Will this population survive or die out eventually?

    How could I use differential equation to model this problem?

    If I let P(t) be the population changes with time t (in days),
    I know that without outside factor, P(t) = 100*2(t/10)
    and with that outside factor, P(t) = P(t-1)*0.92
    and P(0) = 100.

    Then is it true to say that dP/dt = P*2(t/10) ?
    Or how could I set up the equation of dP/dt ?
    Hope you could give me some inspiration.

    Thank you
    Ken

  2. #2
    Elite Member
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    Quote Originally Posted by ken0605040 View Post
    Hi everyone,

    I got a question on a math quest book.

    The questions said in a lake, the population of bacteria will grow at a rate that is proportional to their current population and that in the absence of any outside factors the population will double every 10 days. It is estimates that, on average, (it is an outside factor), on any given day 8% of them will die. The current population is 100.

    Will this population survive or die out eventually?

    How could I use differential equation to model this problem?

    If I let P(t) be the population changes with time t (in days),
    I know that without outside factor, P(t) = 100*2(t/10)
    and with that outside factor, P(t) = P(t-1)*0.92
    and P(0) = 100.

    Then is it true to say that dP/dt = P*2(t/10) ?
    Or how could I set up the equation of dP/dt ?
    Hope you could give me some inspiration.

    Thank you
    Ken
    To get P(t) = 100*2(t/10) , which ODE did you solve?
    “... mathematics is only the art of saying the same thing in different words” - B. Russell

  3. #3
    Should it be dP/dt = P*2(t/10)?

    But it doesn't consider the factor that 8% of population decrease.

  4. #4
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    Quote Originally Posted by ken0605040 View Post
    Should it be dP/dt = P*2(t/10)?

    No....

    How did you get the function without getting the ODE?


    The initial ODE (w/o) external factor is:

    [tex]\dfrac{dP}{dt} = kP[/tex]

    Now continue.....


    But it doesn't consider the factor that 8% of population decrease.

    You'll have to add the population decrease term into the ODE.

    You cannot solve the problem, if you do not know how to set up the ODE without the external factor.
    ,
    “... mathematics is only the art of saying the same thing in different words” - B. Russell

  5. #5
    Thanks, it makes more sense now.

    If I combine the external factor into ODE, can I say

    dP/dt = kP*0.92^t

    if so, then I can find k by using
    P(0) = 100

    Am I right? But how can I treat 'the population will double every 10 days'?
    Can I say P(10) = 200 ?

    It is still a bit confusing.

  6. #6
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    Quote Originally Posted by ken0605040 View Post
    Thanks, it makes more sense now.

    If I combine the external factor into ODE, can I say

    dP/dt = kP*0.92^t

    if so, then I can find k by using
    P(0) = 100

    Am I right? But how can I treat 'the population will double every 10 days'?
    Can I say P(10) = 200 ?

    It is still a bit confusing.
    Before combining the external effect, calculate the valuue of 'kold' - then write the ODE for the external effect.
    “... mathematics is only the art of saying the same thing in different words” - B. Russell

  7. #7
    dP dt Do you mean after finding Kold by using
    1. dP/dt = kP
    2. P(0) = 100

    then, I should work on the ODE for the external effect(i.e. the 8% decreasing population per day)?
    And the ODE become dP/dt = koldP*0.92^t ?

    Sorry that I am not sure if it is what you mean.


  8. #8
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    Quote Originally Posted by ken0605040 View Post
    dP dt Do you mean after finding Kold by using
    1. dP/dt = kP
    2. P(0) = 100

    yes ..... and P(10) = 200

    then, I should work on the ODE for the external effect(i.e. the 8% decreasing population per day)?
    And the ODE become dP/dt = koldP*0.92^t ?

    No... how are you getting the "^t" term included in ODE?

    Please pay attention to what you are writing!

    Sorry that I am not sure if it is what you mean.

    .
    “... mathematics is only the art of saying the same thing in different words” - B. Russell

  9. #9
    Yes, it is not correct at all.

    I want to ask if the result is something like

    P = P0 e(koldknew*t)

    and by using P(1) = 92 to find knew ?

  10. #10
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    Quote Originally Posted by ken0605040 View Post
    Yes, it is not correct at all.

    I want to ask if the result is something like

    P = P0 e(koldknew*t)

    and by using P(1) = 92 to find knew ?
    N0....

    Before the introduction of external factor:

    [tex]\dfrac{dP}{dt} \ = \ k_{old}*P[/tex]

    After the introduction of external factor:

    [tex]\dfrac{dP}{dt} \ = \ k_{old}*P - 0.08*P \ = \ (k_{old} - 0.08) * P[/tex]

    Now continue......
    “... mathematics is only the art of saying the same thing in different words” - B. Russell

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