Rates of change, compounding rates and exponentiation
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 growi
muroscamsey
Open question
2022-08-16
Rates of change, compounding rates and exponentiationI 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?
Answer & Explanation
Isaias Archer
Beginner2022-08-17Added 11 answers
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
for r. That means
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