Language growth with diffrential equations.

1. Jan 12, 2014

cp255

1. The problem statement, all variables and given/known data
There are now about 3300 different human "language families" in the whole world. Assume that all these are derived from a single original language, and that a language family develops into 1.58 language families every 5860 years. About how long ago was the single original human language spoken?

2. Relevant equations

3. The attempt at a solution

So I figured I could model it with this equation dL/dt = kL where k is a constant equal to 1.58/5860. Solving this equation gives me t = (ln(L) + C) / k.

Then plunging in 3300 at time t=0 gives me C = -ln(3300). Next I plug in L=1 and this gives me a time of 30 thousand years ago. Where am I going wrong?

2. Jan 12, 2014

Dick

You are wrong to assume the constant is k=1.58/5860. Why don't you actually try to figure it out instead of assuming?

3. Jan 12, 2014

cp255

I thought it made sense since each language is generating a new one at the rate of k languages per year. Therefore the total rate of change would just be the product of L and k.

4. Jan 12, 2014

Dick

That much is correct. It's the value of k that's wrong. The solution to that differential equation is $L=C e^{kt}$. If you want 1 language at t=0 and 1.58 at t=5860, I don't think you will find k=1.58/5860.

5. Jan 12, 2014

cp255

Yes but isn't L > 5860 at t = 1 because during the interval 0 < t < 1 other languages would have been generated since the growth is continuous.

6. Jan 12, 2014

haruspex

You seem to have L and t swapped there. Did you mean "isn't L > 1.58 at t = 5860"?
Well, yes, it would be if k were 1.58/5860. That's exactly the point Dick is making.
You know $L = Ce^{kt}$ for some C and k; you know L at t = 0; you know L at t = 5860. So calculate C and k.

7. Jan 12, 2014

Dick

Yes, the growth is continuous. But in 5860 years you generate 1.58 languages nonetheless. That's what you given. The k must fit that. k=1.58/5860 is too large. Figure out what it should be.

8. Jan 13, 2014

cp255

I will try it. The way I read the question is that each language generates a new one at the rate of 1.58 languages per 5860 years.

9. Jan 13, 2014

haruspex

Not 1.58 additional ones. Each one becomes 1.58 after 5860 years. So after 2930 years each one becomes √1.58, etc. As Dick wrote, if you use k = 1.58/5860 in ekt you'll get too many languages. After 5860 years you'll have e1.58.

10. Jan 13, 2014

epenguin

I don't know whether this is a exercise math of exponentials or just a crude and practical calculations. If the latter, starting with 1, the number increases by a factor of 1.58 n times to get the number 3300. So work out n that gives you 3300 starting from 1.

Then the time that will have taken is n $\times$ 5860.