Posted on 12/05/2005 3:00:51 PM PST by dawn53
The brithrate is 0.5 antelopes/year/antelope. The death rate is 0.1/year. The two cannot be equal if the population is stable.
The avg lifespan is 10 years. Since there are 5 born per decade, 5 die per decade. The population just didn't pop up and start all in phase, it contains animals of all ages.
If the animals have an average lifespan of ten years, then on average 1/10 of the population will die off each year. If the population is stable, that means an average number of yearly births also equal to 1/10 of the population. It doesn't matter--at least not over a thousand-year timespan--whether the average lifespan of ten years is a result of every animal dying precisely on its tenth birthday, or if it's a result of 99% of animals dying on their ninth birthday while 1% make it to age 109.
Just a thought: no mathematics is required in this problem, just some arithmetic. There is not even a differential equation involved, although it would be more interesting if there were.
Of course, if you remove the stable population restriction the problem can be made to make sense, but that would involve...gasp...variables.
And, this seems to model asexual reproduction. Kinda like the Fibonacci rabbits.
12?
woohooo!!
me and you agreed :)
Also, the 4/5 births death assumption would mean they aren't included in the average 10-year lifespan.
Right.
" The death rate is 0.1/year. The two cannot be equal if the population is stable.
You're given the average lifespan, not the deathrate. You find the deathrate by noting that a stable population is a given and that requires the deathrate = birthrate.
deathrate = 1/(average lifespan).
If there are five antelopes born and five antelopes dying per decade, the average lifespan is not ten years--it is two years.
Consider: in a million years (number chosen large so as to render phasing effects trivial) there will have been 12,000,000,000 antelopes that lived, 24,000 at a time, a total of 24,000,000,000 antelope-years. It doesn't matter whether some animals lived for millenia while others lasted only minutes. The total number of antelope-years accummulated will be 24,000,000,000 (since a herd of 24,000 antelope will accumullate 24,000 antelope-years per year) and the total number of antelope will be 12,000,000,000 (birth rate of 5 antelope per antelope per decade). Thus an average of two antelope-years per antelope, or (more simply), two years.
BTW, my 1980's biology textbook considered the lifespan of bacteria to be infinite, since once a bacterium has split it can only be said to "die" if both offspring have done so. My personal take was (and remains) that average lifespan computations don't require any knowledge of individual lifetimes; counting the number of dying organisms per unit population will give an answer that is no less meaningful for bacteria than for antelopes (even though a few organisms might survive forever, they form such a small portion of the total number that their effect on the average is limited).
Here's the funny part, "the kid"(as he is affectionately know around here) is 17 and about to get his AA (we did dual credit instead of high school.)
He didn't ask me for help in his Calc w/Analytical Geometry classes because, well, frankly, I couldn't have helped him. But he did fine.
So here's, what should be a simple arithmetic problem, but because the problem is so poorly designed, he comes to me with a "Huh, what is this guy talking about?" and we get in a big discussion about "stable populations" and as has been said, male/female ratio, etc.
I'm sure the prof didn't think he was throwing out a controversial problem, it's only homework, that he probably won't spend much time looking over.
I'm just glad he doesn't come to me for help much. My brain is definitely suffering from "mental pause."
Thanks to everyone for their input. When we were having the discussion, I told him, I'll post it on FR and I know you'll get plenty of opinions.
LOL
hey I KNEW taking Calculus in college would work for me eventually......
BTW I got an A last quarter and I am bragging about it :-)
yeah and add to that it's a word problem.
I don't know about everyone else, but those have always given me fits......
Easy, if the species in question normally gives birth to more than one infant at a time. I don't know about the hyothetical antelope in class question, however, since there are multiple species in Africa of antelope-like animals.
Corrections requested if I have calculated incorrectly.
Standard practice when calculating birth rates is to specify it for some number of unsexed individuals; depending upon the male/female ratio, the female birthrate will be about twice that of the population as a whole while the male birthrate will be zero. If, within the antelopes, males and females are represented in equal numbers at all life stages, each female will give birth to an antelope per year, of which half will be males. Or an average of one female every two years.
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