# Homework Help: 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.