Continuous exponential growth and misleading rate terminology

Question

I'm learning about continuous growth and looking at examples of Continuously Compounded Interest in finance and Uninhibited Growth in biology. While I've gotten a handle on the math, I'm finding some of the terminology counterintuitive. The best way to explain would be through an example.

A culture of cells is grown in a laboratory. The initial population is 12,000 cells. The number of cells, $N$, in thousands, after $t$ days is, $N(t)=12e^{0.86t}$, which we can interpret as an $86\%$ daily growth rate for the cells.

I understand the mechanism by which $0.86$ affects the growth rate, but it seems a misnomer to say there's an "$86\%$ daily growth rate" for the cells, as that makes it sound like the population will grow by $86\%$ in a day, when it actually grows by about $136\%$ since the growth is occurring continuously.

Is it just that we have to sacrifice accuracy for succinctness?

Look at the last couple of pages of my handout on exponential growth that I recently posted here (labeled as geom-growth.pdf). — Dave L. Renfro, Mar 27 '13 at 20:33

score 1 · Accepted Answer · answered Mar 26 '13 at 01:23

1

The instantaneous growth rate is $0.86$ per day in that $N(t)$ is the solution to $\frac {dN}{dt}=0.86N$. You are correct that the compounding makes the increase in one day $1.36$

answered Mar 26 '13 at 01:23

Ross Millikan

374,822

Ok, I guess I just have to get used to the term. I suppose I also wanted to make sure I was understanding it correctly. Thanks! – ivan Mar 26 '13 at 12:49

score 0 · Answer 2 · answered Jan 23 '14 at 07:41

0

A rate of 86% that is compounded continuously for t days is the input for your N(t) function. I interpret the rate in English this way.

answered Jan 23 '14 at 07:41

kumar k

33

Continuous exponential growth and misleading rate terminology

2 Answers2