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Math Help - Exponential growth problem

  1. #1
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    Exponential growth problem

    Guys,

    I can't seem to solve this problem. Maybe you guys can help me out?
    Meanwhile, I will be looking for my old math books

    "Consider a population that begins growing exponentially at a rate of 15% per year and then follows a logistic growth pattern. If the carrying capacity is 100 million, find the actual growth rate when the population is 10 million, 50 million and 90 million."

    Thanks a bunch..
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  2. #2
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    It's a little tough to know what to say, since you posted in this generic category. Had you posted in Algebra, there would be a good answer. Had you posted in Differential Equations, there would be simpler answer. The nature of the question rather leans toward the differential equation formulation:

    \frac{dP}{dt}\;=\;r*P\left(1-\frac{P}{K}\right)

    The Initial Growth Rate is r = 0.15
    The Carying Capacity is K = 100 (in millions)

    You have only to substitute your given population values and calculate the actual growth at that level. Calculating the Growth Rate is a matter of dividing by the population again. At P = 80, I get 3%.
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  3. #3
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    Quote Originally Posted by TKHunny View Post
    It's a little tough to know what to say, since you posted in this generic category. Had you posted in Algebra, there would be a good answer. Had you posted in Differential Equations, there would be simpler answer. The nature of the question rather leans toward the differential equation formulation:

    \frac{dP}{dt}\;=\;r*P\left(1-\frac{P}{K}\right)

    The Initial Growth Rate is r = 0.15
    The Carying Capacity is K = 100 (in millions)

    You have only to substitute your given population values and calculate the actual growth at that level. Calculating the Growth Rate is a matter of dividing by the population again. At P = 80, I get 3%.
    thank you.
    Last edited by c47v3770; August 8th 2008 at 06:30 PM.
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  4. #4
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    Thanks for the answer!
    So P= population right?

    So when I substitute to find the actual growth for 10 million, I get:

    0.15 x 10.000.000(1-10.000.000/100.000.000)

    Am I correct?

    and then you said that "Calculating the Growth Rate is a matter of dividing by the population again."

    So I divide the result from the equation above by the population (10.000.000) ???

    I appreciate your help
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  5. #5
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    You have it. The notation is a little awkward because the definition is not quite in the terms you want. Rather then multiply by the population, just to divide by it again, you can just leave it out if all you want is the rate.
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  6. #6
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    Quote Originally Posted by TKHunny View Post
    You have it. The notation is a little awkward because the definition is not quite in the terms you want. Rather then multiply by the population, just to divide by it again, you can just leave it out if all you want is the rate.
    Thank you for the reply.

    Could you please reword this:

    "Rather then multiply by the population, just to divide by it again, you can just leave it out if all you want is the rate."

    I don't really understand what you said. Did you mean to say that I don't have to multiply and divide if all I want is the rate?

    Thank you!
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  7. #7
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    The Definitions

    \frac{dP}{dt} is the Growth over some unit of time.

    \frac{\frac{dP}{dt}}{P} is the Growth Rate over some unit of time.

    The Algebra

    \frac{dP}{dt}\;=\;rP\left(1-\frac{P}{K}\right) is the Growth over some unit of time.

    \frac{\frac{dP}{dt}}{P}\;=\;\frac{rP(1-\frac{P}{K})}{P}\;=\;r\left(1-\frac{P}{K}\right) is the Growth Rate over some unit of time.

    In the last statement, we see that it is not necessary to calculate the Growth, if all you need is the Growth Rate. The algebra has provided us with a shorter route to calculate the Growth Rate.
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  8. #8
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    Quote Originally Posted by TKHunny View Post
    The Definitions

    \frac{dP}{dt} is the Growth over some unit of time.

    \frac{\frac{dP}{dt}}{P} is the Growth Rate over some unit of time.

    The Algebra

    \frac{dP}{dt}\;=\;rP\left(1-\frac{P}{K}\right) is the Growth over some unit of time.

    \frac{\frac{dP}{dt}}{P}\;=\;\frac{rP(1-\frac{P}{K})}{P}\;=\;r\left(1-\frac{P}{K}\right) is the Growth Rate over some unit of time.

    In the last statement, we see that it is not necessary to calculate the Growth, if all you need is the Growth Rate. The algebra has provided us with a shorter route to calculate the Growth Rate.
    Awesome!
    So I just use the formula:

    \frac{\frac{dP}{dt}}{P}\;=\;\frac{rP(1-\frac{P}{K})}{P}\;=\;r\left(1-\frac{P}{K}\right)

    and substitute?

    Sorry if I am asking too many questions..
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  9. #9
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    Go forward! No need to step gingerly. You have thought it through. You have done the algebra. Abandon anything except confidence.
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  10. #10
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    Thanks so much!

    Quick question: do I also have to substitute for

    dP
    ____
    dt
    ______ =
    P

    You have been of great help
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  11. #11
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    Please don't. That's what the problem wants. Once you have it, leave it alone.

    I warned you up front that the notation was a little awkward. You are probaly happy with something like x = 2. Yea! I'm done! I found the value of x.

    When we seed (dP/dt)/P and we get (dP/dt)/P = 2.4%, it's not quite so obvious that we are done.
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