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MathSofiya
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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.
 one year ago
 one year ago
MathSofiya Group Title
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.
 one year ago
 one year ago

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TuringTest Group TitleBest ResponseYou've already chosen the best response.1
this terminology is the only thing confusing me here...
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
I guess I don't know what a "logistic" DE is
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.1
dw:1343418531683:dw
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
ok... that makes sense
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.1
something like population growth or bacteria growth ... etc etc that depends on initial population
 one year ago

MathSofiya Group TitleBest ResponseYou've already chosen the best response.1
\[y=6000e^{0.0015t}\]
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.1
dw:1343508816527:dw
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.1
these kids of equation, differentiate once .... remove these termsdw:1343509003699:dw
 one year ago

dpaInc Group TitleBest ResponseYou've already chosen the best response.0
are those logistic functions? the 6000 is not your initial population, it's your carrying capacity.... the differential equation is: \(\large \frac{dy}{dt}=ky(6000y) \)
 one year ago

MathSofiya Group TitleBest ResponseYou've already chosen the best response.1
What's the thought process...how could I come up with this on my own
 one year ago

lgbasallote Group TitleBest ResponseYou've already chosen the best response.0
...so what's this difference from the usual differential equation for exponential growth?
 one year ago

MathSofiya Group TitleBest ResponseYou've already chosen the best response.1
so we start with \[y=y_0e^kt\] and we take the derivative of this to find a rate of change?
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
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\[y=y_0e^{kt}\]
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
I just don't know all the terminology above lol, but I'm familiar with the problems
 one year ago

MathSofiya Group TitleBest ResponseYou've already chosen the best response.1
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?
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
\[e^{kt+C}=e^{kt}\cdot e^C\]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\[e^{kt+C_1}=C_2e^{kt}\]
 one year ago

MathSofiya Group TitleBest ResponseYou've already chosen the best response.1
Oh I see. Now that I have this equation \[y(t)=Ce^{kt}\] why am I taking the derivative again?
 one year ago

MathSofiya Group TitleBest ResponseYou've already chosen the best response.1
or do I just plug in numbers at this point?
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
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
 one year ago

MathSofiya Group TitleBest ResponseYou've already chosen the best response.1
@dpalnc statement looks legit...(don't know why) \[\large \frac{dy}{dt}=ky(6000y)\]
 one year ago

MathSofiya Group TitleBest ResponseYou've already chosen the best response.1
check this out: \[\frac{dP}{dt}=kP(1\frac{P}{K})\]
 one year ago

MathSofiya Group TitleBest ResponseYou've already chosen the best response.1
Now I'm really confused
 one year ago

MathSofiya Group TitleBest ResponseYou've already chosen the best response.1
I am soo sorry! My book calls the logistic equation little k and carrying capacity big K....
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
I'm just going to say what I'm sure of, which is the initial derivation
 one year ago

MathSofiya Group TitleBest ResponseYou've already chosen the best response.1
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.
 one year ago

MathSofiya Group TitleBest ResponseYou've already chosen the best response.1
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!
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
I guess that's what I'd do as well :p
 one year ago

MathSofiya Group TitleBest ResponseYou've already chosen the best response.1
Sounds like a plan :P Thanks Turing
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
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 :)
 one year ago
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