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Homework Question - Fundamental Theorem of Calc Example

  1. Apr 28, 2015 #1
    1. The problem statement, all variables and given/known data
    Hi, I've been working through a practice problem for which I used the fundamental theorem of calculus, or one of its corollaries.

    The setup is a population changing over time. The population, P(t) at t = 0 is 6 billion. The limiting population as t goes to infinity is given to be 30 billion. The rate of change of the population, [itex] P'(t) = \frac{Ae^t }{ (0.02A + e^t)^2} [/itex] for [itex] t \geq 0 [/itex] (in billions).

    I am asked to find the the time, t, at which [itex] P(t) = [/itex] 10 billion. Throughout, the unit associated with these expressions is "billions of people" in order to express the quantities more succinctly. That is not the issue.

    I have already figured out the solution. It agrees with the solution given by the solutions manual I'm working with. However, there is a step in the solutions manual done differently than how I did it, and I'd like to understand the reasoning.

    2. Relevant equations

    Let [itex]P(t)[/itex] represent the number of individuals in a population at time [itex] t \geq 0 [/itex] (in billions).

    I am given:

    [itex] P'(t) = \frac{Ae^t }{ (0.02A + e^t)^2} [/itex] for [itex] t \geq 0 [/itex] (in billions)

    [itex] \lim_{ t \to \infty} P(t) = 30 [/itex] (billions).

    [itex] P(0) = 6 [/itex] (billions).

    3. The attempt at a solution

    I have the solution for the time at which the population will be 10 billion. I am concerned with understanding an alternate reasoning I found in my solutions manual.

    First I set up an expression using FTOC (or one of the corollaries)

    [tex] P(s) - P(0) = \int_{0}^{s} P'(t) dt[/tex].

    Plugging in the expression for [itex]P'(t)[/itex] and integrating using u-substitution, I was able to get an expression in terms of [itex]s[/itex] in the RHS. Using the substitution [itex] u = 0.02A + e^t [/itex], I get

    [itex] P(s) - P(0) = \frac{A}{0.02A + 1} - \frac{A}{0.02A + e^s} [/itex] (Eq'n ***)

    Using the given information about [itex]P(t)[/itex], I take the limit as t goes to infinity of both sides. I am left with an expression I can use to solve for the constant A. I get a value of A = 46.15.

    Here is where my question lies.

    At this point one can rearrange the (***) equation (above) and get an expression for [itex] P(s) , s \geq 0 [/itex]. One can set it to 10 and find the correct answer for the time. However, this is a bit tedious and messy the way I did. Instead the solution manual gives this expression:


    [tex] P(s) = 30 - \frac{46.15}{0.923 + e^s} [/tex] ... "

    The authors of the solution manual then set this expression to 10 and get the right answer. Where is this expression for P(s) coming from? Is there some corollary I'm missing?
  2. jcsd
  3. Apr 28, 2015 #2


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    Staff: Mentor

    Looks like the same approach. If you calculate ##\frac{A}{0.02 A + 1} + P(0)##, you get 30.
    You can save the step of calculating the value of this fraction, I guess, as you know the result of the expression has to be 30.
  4. Apr 28, 2015 #3
    I don't know how I missed this. Thanks.
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