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MathSofiya

  • 3 years ago

Suppose that a population grows according to a logistic model with carrying capacity 6000 and k=0.0015 per year. Write a logistic differential equation for these data.

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  1. TuringTest
    • 3 years ago
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    this terminology is the only thing confusing me here...

  2. TuringTest
    • 3 years ago
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    I guess I don't know what a "logistic" DE is

  3. experimentX
    • 3 years ago
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    |dw:1343418531683:dw|

  4. TuringTest
    • 3 years ago
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    ok... that makes sense

  5. experimentX
    • 3 years ago
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    something like population growth or bacteria growth ... etc etc that depends on initial population

  6. MathSofiya
    • 3 years ago
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    \[y=6000e^{0.0015t}\]

  7. MathSofiya
    • 3 years ago
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    ?

  8. experimentX
    • 3 years ago
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    |dw:1343508816527:dw|

  9. experimentX
    • 3 years ago
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    these kids of equation, differentiate once .... remove these terms|dw:1343509003699:dw|

  10. dpaInc
    • 3 years ago
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    are those logistic functions? the 6000 is not your initial population, it's your carrying capacity.... the differential equation is: \(\large \frac{dy}{dt}=ky(6000-y) \)

  11. MathSofiya
    • 3 years ago
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    What's the thought process...how could I come up with this on my own

  12. lgbasallote
    • 3 years ago
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    ...so what's this difference from the usual differential equation for exponential growth?

  13. MathSofiya
    • 3 years ago
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    so we start with \[y=y_0e^kt\] and we take the derivative of this to find a rate of change?

  14. TuringTest
    • 3 years ago
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    no, start with the idea that the current rate of change is proportional to the current population\[\frac{dy}{dt}\propto y=ky\]separate the variables\[\frac{dy}y=kdt\]integrate both sides:\[\ln y=kt+C\]\[y(t)=Ce^{kt}\]by plugging in t=0 we find that C the original population\[y(0)=Ce^0=C\implies C=y_0\]so we get\[y=y_0e^{kt}\]

  15. TuringTest
    • 3 years ago
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    I just don't know all the terminology above lol, but I'm familiar with the problems

  16. MathSofiya
    • 3 years ago
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    how do we go from \[\ln y=kt+C\] to \[y(t)=Ce^{kt}\] I presume it was doing e^.... on both sides but why is the C multiplied to e?

  17. TuringTest
    • 3 years ago
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    \[e^{kt+C}=e^{kt}\cdot e^C\]and \(e^C\) is just another constant, so I just called it C again. I maybe should have numbered them for the sake of distinction, but that gets tedious in higher math\[e^{kt+C_1}=C_2e^{kt}\]

  18. MathSofiya
    • 3 years ago
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    Oh I see. Now that I have this equation \[y(t)=Ce^{kt}\] why am I taking the derivative again?

  19. MathSofiya
    • 3 years ago
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    or do I just plug in numbers at this point?

  20. TuringTest
    • 3 years ago
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    just plug it in usually, but carrying capacity... not sure of that is at time t=0 sounds like a maximum to me again the terminology is messing me up

  21. MathSofiya
    • 3 years ago
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    @dpalnc statement looks legit...(don't know why) \[\large \frac{dy}{dt}=ky(6000-y)\]

  22. MathSofiya
    • 3 years ago
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    check this out: \[\frac{dP}{dt}=kP(1-\frac{P}{K})\]

  23. MathSofiya
    • 3 years ago
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    Now I'm really confused

  24. MathSofiya
    • 3 years ago
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    I am soo sorry! My book calls the logistic equation little k and carrying capacity big K....

  25. TuringTest
    • 3 years ago
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    I'm just going to say what I'm sure of, which is the initial derivation

  26. MathSofiya
    • 3 years ago
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    I think I'm kinda done with this problem LOL. I suck at word problems. But your steps in the initial derivative makes sense though.

  27. MathSofiya
    • 3 years ago
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    It seems like my book just gave me this equation \[\frac{dP}{dt}=kP(1-\frac{P}{K})\] and just wants me to plug and chug numbers. ....fine with me!

  28. TuringTest
    • 3 years ago
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    I guess that's what I'd do as well :p

  29. MathSofiya
    • 3 years ago
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    Sounds like a plan :P Thanks Turing

  30. TuringTest
    • 3 years ago
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    that equation makes sense too because when P=K, dP/dt=0 i.e. when the population = max capacity it stops changing reasonable to me :)

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