Population growth

IrishKnight

The doubling time of the population of a country is estimated at 100 years, from some fixed point in time. Assuming a constant growth rate, it is estimated that after 1000 years from this time the country’s population will be what?


At first I thought it to be P(t)=(2^t)(P(o)) However that is not one of the answers that I can pick from, them being 10, 5, 32, 16, 100.

Any pointers?

EL_Loco

The answer is 10, but I had to draw a graph

went from the 1 on the y axis at 0 on the x, to 2 on y and 100 on x, ( no point is starting at 0 becuase you can't double nothing ) and projected the line out from there, went to the year 1100, because it says 1000 years from this point (after the initial 100 years).

hope that's of some use.

IrishKnight

Won't it be a exp growth curve?

EL_Loco

like this?

year	population
0	 1
100	2
200	4
300	8
400	16
500	32
600	64
700	128
800	256
900	512
1000	1024

does constant growth rate mean it'll double every 100 years, or the rate it took to double from 1 to 2 after 100 is the rate (which is what I graphed). man, what a straight forward, yet fooked up question.

Michael Collins

You are not given a starting population so the answer can't be a number. What value are you actually looking for? Is it the factor of increase? If it is you should be right with your formula:

i.e. population doubles every 100 years, so after 1000 years it has doubled
1000/100 = 10 times. Hence the factor of increase is 2^10 = 1024.

The miniumum population possible is 0 (corresponds to nobody there at t = 0)
The next smallest value is 1024 (mathematically anyway, for pop. = 1 at t = 0)

The last value has leapfroged all your allowable answers, so you must be looking for something else?

Pherekydes

As mentioned earlier, population growth is exponential. That was the first thing I thought of when I saw the thread title. But the question is worded confusingly, I think.

Take a look at this eqn. Does it make any sense?

p new = p0 * e^tx

If p new = 2p0 then

tx = ln 2 (= approx 0.7)

Now look at this:



Subbing 100 in place of the doubling time, we get a 0.7% per annum growth rate. Which I got for tx above.

If tx = .7 and t = 100 then x = .007

Subbing 1000 for t, we get

p new = p0 * e ^ 6.93

and e ^ 6.93 is 1024, which several people have suggested as the answer. This figure is, of course, the multiple of the original population, which is what I suspect the original question to have asked.

Sense or NON sense?

Will I get me coat?

Edited to add references:

Lecture on Pop growth

Population modeling

EL_Loco

at this stage I'd like to see the answer, and reasoning behind it. if that's obtainable. OP was it a college/school question?

IrishKnight

I would like to see the reason behind it too! It was on a first year calc sample mid term. I have yet to get an answer but I will email the good Dr. for an answer.

Thanks for all the replys by the by.

IrishKnight

Ok just go an email back say that I was right in thinking that is was 1024 times bigger...