anonymous
  • anonymous
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.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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TuringTest
  • TuringTest
this terminology is the only thing confusing me here...
TuringTest
  • TuringTest
I guess I don't know what a "logistic" DE is
experimentX
  • experimentX
|dw:1343418531683:dw|

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TuringTest
  • TuringTest
ok... that makes sense
experimentX
  • experimentX
something like population growth or bacteria growth ... etc etc that depends on initial population
anonymous
  • anonymous
\[y=6000e^{0.0015t}\]
anonymous
  • anonymous
?
experimentX
  • experimentX
|dw:1343508816527:dw|
experimentX
  • experimentX
these kids of equation, differentiate once .... remove these terms|dw:1343509003699:dw|
anonymous
  • anonymous
are those logistic functions? the 6000 is not your initial population, it's your carrying capacity.... the differential equation is: \(\large \frac{dy}{dt}=ky(6000-y) \)
anonymous
  • anonymous
What's the thought process...how could I come up with this on my own
lgbasallote
  • lgbasallote
...so what's this difference from the usual differential equation for exponential growth?
anonymous
  • anonymous
so we start with \[y=y_0e^kt\] and we take the derivative of this to find a rate of change?
TuringTest
  • TuringTest
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\[y=y_0e^{kt}\]
TuringTest
  • TuringTest
I just don't know all the terminology above lol, but I'm familiar with the problems
anonymous
  • anonymous
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?
TuringTest
  • TuringTest
\[e^{kt+C}=e^{kt}\cdot e^C\]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\[e^{kt+C_1}=C_2e^{kt}\]
anonymous
  • anonymous
Oh I see. Now that I have this equation \[y(t)=Ce^{kt}\] why am I taking the derivative again?
anonymous
  • anonymous
or do I just plug in numbers at this point?
TuringTest
  • TuringTest
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
anonymous
  • anonymous
@dpalnc statement looks legit...(don't know why) \[\large \frac{dy}{dt}=ky(6000-y)\]
anonymous
  • anonymous
check this out: \[\frac{dP}{dt}=kP(1-\frac{P}{K})\]
anonymous
  • anonymous
Now I'm really confused
anonymous
  • anonymous
I am soo sorry! My book calls the logistic equation little k and carrying capacity big K....
TuringTest
  • TuringTest
I'm just going to say what I'm sure of, which is the initial derivation
anonymous
  • anonymous
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.
anonymous
  • anonymous
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!
TuringTest
  • TuringTest
I guess that's what I'd do as well :p
anonymous
  • anonymous
Sounds like a plan :P Thanks Turing
TuringTest
  • TuringTest
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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