4
$\begingroup$

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:

$P=P_0e^{rt}$

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):

$1000*e^{1*0.1}=1105.17$

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?

Thanks,

Billy.

$\endgroup$
3
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$
1
$\begingroup$

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:

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

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.