I have a very (apologies if stupidly) simple question about rates of change that has been bugging me for some time. I can't work out whether it relates to my misunderstanding what a rate of change is, to my misapplying the method for calculating a rate of change or something else. I'm hoping somebody on here can help.

For how I define a rate of change, take as an example a population of 1000 items (e.g. bacteria). I observe this population and after an hour I count the size of the population and see that it has increased by 10% (to 1100). I might hypothesise that the population is growing at the rate of 10% per hour, and if, an hour later, I see that it has grown by 10% again (to 1,210) then I might decide to conclude that it is growing at 10% per hour.

So, a rate of change of "proportion x per hour" means "after one hour the population will have changed by proportion x". If, after 1 hour, my population of bacteria was not 1,100, and if not 1,210 after 2 hours, that would mean that the rate of change was not 10% per hour.

First question: Is this a fair definition of a rate of change?

So far so good and it's easy to calculate the population after any given time using a compound interest-type formula.

But whenever I read about continuous change something odd seems to happen. Given that "grows at the rate of 10% per hour" means (i.e. is just another way of saying) "after 1 hour the original population will have increased by 10%", why do textbooks state that continuous change should be measured by the formula:


And then give the rate of change in a form where this seems to give the wrong answer (i.e. without adjusting it to account for the continuously compounded growth)? I've seen many texts and courses where 10% per day continuous growth is calculated as (for my above example, after 1 day):


This contradicts the definition of a rate of change expressed as "x per unit of time" stated above. If I was observing a population of 1000 bacteria and observed it grow to a population of 1105 after 1 hour I should surely conclude that it was growing at the rate of 10.5% per hour.

I can get the idea of a continuous rate just fine, and it's easy to produce a continuous rate of change that equates to a rate of 10% per day as defined above (that's just ln 1.1). But I struggle to see how a rate of change that means a population grows by 10.5% in an hours means it is growing at 10% per hour. That's like saying if I lend you money at 1% interest per month I'd be charging you 12% per year.

So what's wrong here? Have I got the wrong end of the stick with my definition of a rate of change, would most people interpret a population increase of 10.5% in an hour as a 10% per hour growth rate, or is something else amiss?




2 Answers 2


The short answer to your question is that the $10$ percent growth you observed after one hour was the result of continuous compounding (growth) at some rate $r$ throughout the hour. To find that $r$ you solve $$ e^{r \times 1} = 1.1 $$ for $r$. That means $$ r = \ln 1.1 \approx 0.095. $$ That's a little less than $0.1$ because of the compounding.

When you see the growth rate reported as $10$ percent per hour it is indeed a little ambiguous. The writer may mean that the population is given by $$ P_0e^ {0.1t} $$ or by $$ P_0e^ {0.095t}. $$ You need the context to disambiguate.


One uses a reference time unit (e.g., one year for money, one hour for bacteria) to define compounding frequency for wealth (gain) accumulation from time $t$ to time $T$ (observed wealth over different pairs of times $t$ and $T$ can change and it will imply different rates of growth).

A $n$-times-per-time unit compounded rate is a constant rate $y^n(t,T)$ (referred to the time unit) at which one grows initial quantity $1$ (dollar or bacterium) at time $t$ to produce wealth (gain) $w(t,T)$ by time $T$, where

$$w(t,T) = \left(1 + \frac{y^n(t,T)}{n} \right)^{n\cdot \tau(t,T)} $$ with time difference $\tau(t,T)$ expressed in reference time units (e.g., if $t,T$ are dates and reference time unit is one year, the time difference is $(T-t)/365$ years).

For $n=1$, we get the familiar $y^1(t,T)$ with wealth $$w(t,T) = \left(1 +y^1(t,T) \right)^{\tau(t,T)} $$

For $n\rightarrow \infty$, we get the continuously-compounded rate $y^\infty(t,T)$ with wealth

$$w(t,T) = \lim_{m\rightarrow \infty}\left(1 + \frac{y^\infty(t,T)}{m} \right)^{m\cdot \tau(t,T)} = \mathrm{e}^{y^\infty(t,T) \tau(t,T)} $$

There is also the simply-compounded rate $y^0(t,T)$ for which accruing is proportional with time:

$$ w(t,T) = 1 +y^0(t,T) \tau(t,T) $$


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