## 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. 2 years ago 2 years ago

1. TuringTest Group Title

this terminology is the only thing confusing me here...

2. TuringTest Group Title

I guess I don't know what a "logistic" DE is

3. experimentX Group Title

|dw:1343418531683:dw|

4. TuringTest Group Title

ok... that makes sense

5. experimentX Group Title

something like population growth or bacteria growth ... etc etc that depends on initial population

6. MathSofiya Group Title

$y=6000e^{0.0015t}$

7. MathSofiya Group Title

?

8. experimentX Group Title

|dw:1343508816527:dw|

9. experimentX Group Title

these kids of equation, differentiate once .... remove these terms|dw:1343509003699:dw|

10. dpaInc Group Title

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 Group Title

What's the thought process...how could I come up with this on my own

12. lgbasallote Group Title

...so what's this difference from the usual differential equation for exponential growth?

13. MathSofiya Group Title

so we start with $y=y_0e^kt$ and we take the derivative of this to find a rate of change?

14. TuringTest Group Title

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 Group Title

I just don't know all the terminology above lol, but I'm familiar with the problems

16. MathSofiya Group Title

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 Group Title

$e^{kt+C}=e^{kt}\cdot e^C$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 Group Title

Oh I see. Now that I have this equation $y(t)=Ce^{kt}$ why am I taking the derivative again?

19. MathSofiya Group Title

or do I just plug in numbers at this point?

20. TuringTest Group Title

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 Group Title

@dpalnc statement looks legit...(don't know why) $\large \frac{dy}{dt}=ky(6000-y)$

22. MathSofiya Group Title

check this out: $\frac{dP}{dt}=kP(1-\frac{P}{K})$

23. MathSofiya Group Title

Now I'm really confused

24. MathSofiya Group Title

I am soo sorry! My book calls the logistic equation little k and carrying capacity big K....

25. TuringTest Group Title

I'm just going to say what I'm sure of, which is the initial derivation

26. MathSofiya Group Title

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 Group Title

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 Group Title

I guess that's what I'd do as well :p

29. MathSofiya Group Title

Sounds like a plan :P Thanks Turing

30. TuringTest Group Title

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 :)