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Exponential growth constant

  1. Dec 29, 2009 #1
    I am learning calculus from a book of mine, and it gave an example problem of exponential growth (as derived from the exponential differential equation of dy/dx = ky to be y=Ce^kt) saying a population is growing at a rate of 2% per year, and says that K, the growth constant, in this case is k = .02. My confusion is, I thought you could set up the equation of this to be Y=A(1.02)^(t) "A" being the initial amount, and here I can see it is also Y=A*e^(.02*t) since k=.02 yet these equations are not the same and I am very confused because if you plug in one year length (t=1) then the initial amount has grown by 1.02 (or increased by 2%) and this is not the case for the second equation. Could someone explain to me whether the book has set k to the wrong value (if an amount grows by x% by some time, the growth constant k is not that percentage) or why these equations are different, or if one is incorrect.

    (The book says an amount is growing by 2% and then says that the growth constant k=.02)
     
  2. jcsd
  3. Dec 30, 2009 #2
    hmm. k should be ln(1.02)
     
  4. Dec 30, 2009 #3
    This is the difference between descrete time recurrence relation and continuous time "recurrence" relation.
    In descrete time Y[n+1]=k*Y[n] you are dealing with sequences, and here the solution is a geometric series with q=k. In descrete time the equation implies a direct relation between the current value and the next value.
    In continuous time the relation y'(t)=ky(t), you are dealing with function. Here the equation implies a relation between the current value of the function and its current rate of change.
    That what sets the difference.

    But by "sampling" you can rearrange the latter equation to take the form of the first equation.
    Let's say we sample our function at intervals of [tex]\delta t[/tex] and we define Y[n] to be [tex]y(t=n\delta t)[/tex]. So the approximate derivative will be

    [tex]\frac{Y[n+1]-Y[n]}{\delta t}[/tex]

    Comparing with kY[n] (as the diff. equation gives) you have:

    [tex]Y[n+1]=(k\delta t + 1)Y[n][/tex]

    Here, as you said yourself, the discrete solution is:

    [tex]Y[n]=Y[0](k\delta t +1)^{n}[/tex]

    Now let's connect this to the continuous case:
    Remember that [tex]n=\frac{t}{\delta t}[/tex]

    So an approximate evaluation of y(t) is:

    [tex]y(t)=y(0)(k\delta t +1)^{\frac{t}{\delta t}}[/tex]

    To get this approximation more and more accurate, we need to sample more frequently, or in other words, have [tex]\delta t[/tex] be as small as possible. We take the limit of the left-havd expression at [tex]\delta t --> 0[/tex], and if you remember the defenition of e, you get back your expected solution:

    [tex]y(t)=y(0)e^{kt}[/tex]
     
  5. Dec 30, 2009 #4
    wow, thanks elibj. I beleive the book was incorrect, the growth constant K refers to the relation between the current value and it's rate of change, which is not the same constant as the relation between current value and next value. Thanks again you guys.
     
  6. Dec 30, 2009 #5
    I know this is off subject, but what do I need to load onto my computer in order to see your nicely formatted latex formulas? I can see the source code but not the resulting image.
     
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