Introduction
Okay this is embarrassing, but I was always confused about finite and instantaneous rate when I was a young grad student. So, let’s say you’re writing a model (let’s say of insects because it gets really depressing talking about people) and you gotta parameterize the daily mortality rate. You scour through some literature and find an experiment that says they found 10% of the individuals died in a day.
For simplicity sake, the daily mortality parameter is called \(\mu\) and the insect population is \(N(t)\) and we’re going to use differential equations show below:
\[\frac{dN(t)}{dt} = -\mu N(t).\]
Without thinking, let’s plug in 0.10 for \(\mu\). But the problem is that this is a finite rate, we have to convert it first!
Continuous versus discrete death
Let’s think about it, when we talk about differential equations we have to realize that they’re continuous! Meaning that while we like to think about like modeling population dynamics in the manner of days (i.e 10 insects die every day), we have to realize that there are time units smaller than days: hours, minutes, seconds. And at these tiny time steps, there is still death! As differential equations are continuous, that means that death is a continuous process.
Okay, you found that the experiment found that 10% of the insects die in a day. But that is a finite rate! You’re not assuming insects are dying continuously throughout that one day. In fact, it’s like they’re waiting to die at the end of each day!
How to convert
So what does that mean? It means you have to take your finite rate and convert it to an instantaneous rate. How do you do that?
Well let’s look at the differential equation here again:
\[ \frac{dN}{dt} = -\mu N(t)\]
If we solve it then it becomes
\[N(t) = N_0 * \exp(-\mu t).\]
Okay so let’s say we have an initial conditions of 100, so \(N_0 = 100\). I want to solve for \(\mu\). So let’s divide both parts of the equation with \(N_0\) and that means it would \(\frac{N(t)}{N_0}\). Okay so if we see that 10% have died in one day that means \(N(1) = 90\) \((100- (100 * 0.10) = 90)\). Substituting the new numbers in and setting \(t = 1\) since we’re looking at one day only,we get:
\[\frac{90}{100} = \exp(-\mu t) = \exp(-\mu).\]
We can take the natural log to get: \(log(\frac{90}{100}) = -\mu\). When we calculate it we get \(\mu = 0.105\). Aha, close to 0.10 but not exactly 0.10? What was the entire point of this exercise? Well, the finite and instantaneous rates actually diverge as the finite mortality rate increases. Say that 70% of individuals die in a day. So okay, rerun the above. \(\mu = -ln(\frac{30}{100}) = 1.20.\) That’s a lot bigger than 0.70! Remember that RATES allow for a value greater than 1.
If you want more information check:
https://influentialpoints.com/Training/finite-and-instantaneous_rates.htm