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Math Help - Logistic Growth

  1. #1
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    Logistic Growth

    q(t) = a / 1 + be^-kt

    where b = a / q(initial) - 1 and k > 0 is constant.

    Assuming a > q(initial) > 0, prove that the function q is everywhere increasing by analyzing its first derivative. Do not use specific values for any of the constants.

    I found the derivative of q(t) to be

    q'(t) = (-a(be^-kt)-bk) / ((1+be^-kt)^2)

    First of all, is that right? And if it is, how do I prove that the function is everywhere increasing without using specific values for any of the constants?
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  2. #2
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    Quote Originally Posted by WahooMan View Post
    q(t) = a / 1 + be^-kt

    where b = a / q(initial) - 1 and k > 0 is constant.

    Assuming a > q(initial) > 0, prove that the function q is everywhere increasing by analyzing its first derivative. Do not use specific values for any of the constants.

    I found the derivative of q(t) to be

    q'(t) = (-a(be^-kt)-bk) / ((1+be^-kt)^2)

    First of all, is that right? And if it is, how do I prove that the function is everywhere increasing without using specific values for any of the constants?
    Your derivative is wrong. Fix it and then use the fact that a> 0, k > 0 and a > q(initial) > 0.
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  3. #3
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    Quote Originally Posted by mr fantastic View Post
    Your derivative is wrong. Fix it and then use the fact that a> 0, k > 0 and a > q(initial) > 0.
    Is it

    q'(t) = (-a(be^-kt)) / ((1+be^-kt)^2)

    ?
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  4. #4
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    Quote Originally Posted by WahooMan View Post
    Is it

    q'(t) = (-a(be^-kt)) / ((1+be^-kt)^2)

    ?
    No.

    Instead of fumbling around in the dark and posting the result, please show all the details of your calculation so that your mistake(s) can be pointed out.
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  5. #5
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    Quote Originally Posted by mr fantastic View Post
    No.

    Instead of fumbling around in the dark and posting the result, please show all the details of your calculation so that your mistake(s) can be pointed out.
    q(t) = a / 1 + be^-kt

    Quotient Rule: (Derivative of Top(Bottom) - Top(Derivative of Bottom)) / ((Bottom)^2)

    Derivative of Top: 0
    Derivative of Bottom: 0 + be^-kt (I think this is where my problem is)

    q'(t) = (0(1+be^-kt)-a(be^-kt)) / ((1+be^-kt)^2)
    = (-a(be^-kt)) / ((1+be^-kt)^2)
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  6. #6
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    Quote Originally Posted by WahooMan View Post
    q(t) = a / 1 + be^-kt

    Quotient Rule: (Derivative of Top(Bottom) - Top(Derivative of Bottom)) / ((Bottom)^2)

    Derivative of Top: 0
    Derivative of Bottom: 0 + be^-kt (I think this is where my problem is)

    q'(t) = (0(1+be^-kt)-a(be^-kt)) / ((1+be^-kt)^2)
    = (-a(be^-kt)) / ((1+be^-kt)^2)
    Just to be clear: Are you saying that you do not know how to differentiate 1 + b e^{-kt} ? If that is the case, your best course of action is to go back to your class notes or textbook and review differentiation of basic exponential functions.
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  7. #7
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    Quote Originally Posted by mr fantastic View Post
    Just to be clear: Are you saying that you do not know how to differentiate 1 + b e^{-kt} ? If that is the case, your best course of action is to go back to your class notes or textbook and review differentiation of basic exponential functions.
    q'(t) = (-a(-bke^-kt)) / ((1+be^-kt)^2)

    That's it, isn't it? I'm feeling pretty good about that one.
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  8. #8
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    Quote Originally Posted by WahooMan View Post
    q'(t) = (-a(-bke^-kt)) / ((1+be^-kt)^2)

    That's it, isn't it? I'm feeling pretty good about that one.
    Yes. And the numerator simplifies to abk e^{-kt}. Now go back and read the second sentence in post #2.
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  9. #9
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    Quote Originally Posted by mr fantastic View Post
    Yes. And the numerator simplifies to abk e^{-kt}. Now go back and read the second sentence in post #2.
    The function will always be positive. Thanks.
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  10. #10
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    Quote Originally Posted by WahooMan View Post
    The function will always be positive. Thanks.
    Do you understand why b > 0 ....?
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