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As many of us might know, the set up of a differential equation relating instantaneous population growth with respect to time to the per capita population growth rate and current population size at an arbitrary time is as follows:

dN/dt = rN

Now, we are told that the constant, r, is the per capita population growth rate. N is the current size of the population at some time. "r" is the "organisms added per reproductive period" divided by the total population at that time. For instance, if 10 rabbits are born per month and the current population is 100 rabbits, then r = 0.1.

Let's consider a case where a population is doubling every reproductive period, modeled by 2^t. we can choose to model 2^t in base e and equivalently we have e^ln(2)t. So 2^t = e^ln(2)t = e^kt where e^kt is a general formula for exponential growth or decay. For doubling, k = ln(2). In differential equation form we have dN/dt = kN , k = ln(2)

You can't get natural log of 2 by taking the number of organisms added to the population and dividing by the current total population. r does not equal k no matter how you set up your growth rate, but the structure of the relationship is the very same differential equation for exponential growth: dN/dt = kN, which we can integrate to a general form such as A*e^kt. Why do we use this "r" contrivance at all? I am not sure I understand this choice. How is r different from the standard exponential growth factor k? What does it help us to understand in modeling the population growth rate?

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  • $\begingroup$ I'm confused. Your first two paragraphs basically make sense (except that I don't understand why you switched notation from r to k, these seem to be exactly equivalent). I don't understand your last paragraph at all; can you try to clarify? Are you confused about the switch between discrete- and continuous-time models? $\endgroup$
    – Ben Bolker
    Commented Aug 24, 2019 at 2:57
  • $\begingroup$ Integrating $\frac{\mathrm{d}N(t)}{\mathrm{d}t}=rN(t)$ gives you $N(t)=N(0)\exp (rt)$. Doubling would mean the ratio of the population at time $t$ to time $0$ is 2; i.e., $2 = \frac{N(t)}{N(0)} = \exp (rt)$. You can see how $r \neq \log (2)$. As pointed out by @BenBolker above, I think you may be cofounding instantaneous, continuous-time growth (differential equation) with incremental, discrete-time growth (difference equation). $\endgroup$
    – Chris Moore
    Commented Oct 23, 2019 at 2:21

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