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AP Cal DE problem.

  1. Apr 18, 2005 #1
    2004 BC 5
    A population is modeled by a function P that satisfies the logistic differential equation

    dP/dt= P/5(1-P/12)

    a) If P(0)=3, what is Limit[P(t), t, infinite]
    If P(0)=20, what is Limit[P(t), t, infinite]

    do i have a way to evaluate the limit without solving the D.E.??

    Because this is the first sub-question on the entire question, i believe it wouldnt be this complicated.
     
  2. jcsd
  3. Apr 18, 2005 #2

    dextercioby

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    U have to integrate the ODE.It's very easy.It's separable.You might integrate it with corresponding limits.

    Daniel.
     
  4. Apr 18, 2005 #3

    OlderDan

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    You don't have to solve the differential equation to do this problem. If P(0) = 3, the initial value of the derivative is positive. P will increase until the derivative is no longer positive. P will approach that value asymptotically. For what value of P is the derivative zero?

    If P(0) = 20, the derivative is initally negative. P will decrease and approach the value where the derivative is zero.

    You only have to solve the equation if you want the shape of the curve P(t).
     
  5. Apr 19, 2005 #4
    So i have to do everystep?
    or i can just set the slope equal to 0 and then solve for P, which is the aysumtope?
     
  6. Apr 19, 2005 #5

    dextercioby

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    How do you know the function does have (presumably horizontal) asymptotes,without solving the ODE?

    Daniel.
     
  7. Apr 20, 2005 #6

    OlderDan

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    If all you need is the asymptote, this will do it.

    You know it has an asymptote at P = 12 because you know the initial derivative, and the sign of the derivative cannot change without changing the sign of the factor (1-P/12). As P approaches 12, dP/dt approaches zero without ever changing sign or reaching zero. If it ever got to P = 12, the derivative would be zero and P would be stuck there.
     
    Last edited: Apr 20, 2005
  8. Apr 20, 2005 #7
    i knew there is an asymtote becaz the way that the question asks and i graph the slope field, but this is not enough for me to do the reasoning for AP test.
    and i am wondering if i can get credit from seting the slope = 0 or i have to solve the DE. (i figure out the way to solve it.... not that easy....)
     
  9. Apr 20, 2005 #8

    OlderDan

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    From what I have seen of the AP, the authors are fond of these qualitative reasoning type problems. You can solve the equation, or you use the slope field concept to justify the arguement that P will approach the limiting value where the slope becomes zero. Then do the simple calculation you have recognized. Set the slope equal to zero and find the corresponding P value.

    You cannot just set the slope equal to zero and be done with it, because there are two values of P that make the slope zero: P = 0 and P = 12. You need the initial condition to get the initial slope that tells you P is headed in the direction of P = 12. The slope field tells you there is no going back. What would happen if the initial value of P were negative? The initial value is an important part of the problem.

    There are other functions that lend themselves to the same analysis. The function that satisfies dP/dt = -aP where a is a positive constant is well known, but you don't need the function to recognize that if P starts out positive it has a negative derivative and is heading toward the value of P that makes the derivative zero. P can never quite get to zero, because the slope gets less negative as it approaches zero, but the slope is never zero or positive. The P = 0 level serves as an attraction of P toward a level that can never be reached.
     
  10. Apr 20, 2005 #9

    HallsofIvy

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    I assume the equation is dP/dt= (P/5)(1-P/12) (notice the parentheses)

    If P(t)= 0 for all t, then the equation is satisfied. If P(t)= 12, for all t, then the equation is satisfied. That is P(t)=0 and P(t)= 12 are both (constant) solutions to the equation. Furthermore, since the solution to this equation, for given initial value, are unique,other solutions cannot cross those. THAT tells us that P(t)= 0 and P(t)= 12 are asysmptotes for other solutions.

    In particular, if P(0)= 3, which is between 0 and 12, 0< P(t)< 12 for all t. Furthermore, it is clear that (P/5)(1- P/12) is positive for all P between 0 and 12 so this solution will have positive derivative for all t. It must increase, asymptotically, to P= 12 as t increases.

    If P(0)= 20> 12, P(t)> 12 for all t. Further, (P/5)(1- P/12)< 0 for all P> 20 so the derivative of this solution is negative for all t. The function decreases, asymptotically, to P= 12 as t increases.
     
  11. Apr 20, 2005 #10

    saltydog

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    Nice! Can I plug Devaney again? You know, systems like these often exhibit catastrophe: A large pond of fish, I fish for them, remove them from the pond, the population adjust, fish some more, adjust. Bring over friends, we really get them. It's a stress but they adjust. You know, it can get to a point where I push it (do lots of fishing), where removing one more fish and the whole population collapses abruptly to extinction. The asymptotes are key. Jumping over one (by changing a critical parameter) causes the system to bifurcate catastrophically. Lots of things like this in nature. Differential equations . . . they rock.

    Edit: In case you're interested Robert Devaney wrote a fundamentally new kind of book on differential equations stressing the "qualitative analysis" of equations as was done so nicely above by others.

    Edit#2: You know, I think I made a mistake above, really you're not crossing the asymptotes but rather as a parameter is changed the actual number of asymptotes change similar to how the roots of a polynomial change as the plot is shifted up or down. This causes the abrupt change in the system. As for the fish, you know I might have some problems showing that. I'm pretty sure about it but if someone calls me on it . . . maybe I should just keep quiet. It's very interesting though.
     
    Last edited: Apr 20, 2005
  12. Apr 20, 2005 #11

    saltydog

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    Alright, I got one and I hope I'm not annoying anyone. Consider the following ODE which (may) represent a fish population in hundreds of fish in a pond:

    [tex]\frac{df}{dt}=-(f-2)^2[/tex]

    Do I have to say what 'f' represents?

    If we start with a fish population of 500, we'll find that the population drops to 200 then stabalizes at that value. See attached plot labeled 'stable'.

    Now, anything above 200, and the population will drop back to the stable 200 value. The second plot shows what happens when the population drops to 199 fish.


    Maybe not the best example, but some vindication anyway. Alright, I'm done.
     

    Attached Files:

    Last edited: Apr 20, 2005
  13. Apr 20, 2005 #12

    HallsofIvy

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    "Paul Blanchard, Robert L. Devaney, and Glen R. Hall"

    I have that book right beside me now. I'm using it in a D.E. course this semester. Good book, really covers non-linear d.e. much more than other books. Perhaps not the best choice for an introductory course!
     
  14. Apr 20, 2005 #13
    good explaination to solve this question without solving the ODE. thx a lot
    i know how to solve this question with slope field and separation of variables. But this part is only worth 1 point... so i believed it should be easier than those methods.
    and the equation itself is dP/dt= (P/5)(1-P/12)
    i usually following the way to type in mathematica and calculator... sorrie for confusing ppl
     
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