# Bacteria Growth without integration

1. Nov 5, 2009

### Mirole

A bacteria culture initially contains P(o) cells and grows at the rate dP/dt = kP where k is a growth constant. After an hour the population has doubled.

(a) Determine an expression for the number of bacteria present after t hours.

(b) Computer the number of bacteria present, and the rate of growth, after 2 hours.

(c) After how many hours will the population be 10P(o)
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For (a), I could just seperate the differential equation dP/dt= kP as dp/P= kdt and then integrate both sides. For, (b) I would just set t = 2. And, for (c), P(t)= 10P(0).

I'm supposed to do this without using integration, which I have no idea how, any ideas?

Ok, so for (a), I did:

2P(o) = P(o)e^(k*1)
ln2 = e^(k*1)ln
ln(2) = k

So, P = P(o)e^(ln(2)*t)

Did I do it right?

2. Nov 5, 2009

### HallsofIvy

Staff Emeritus
If you know that this must be an exponential, then, yes, that works. Of course, we know that it is an exponential because we have integrated before!

Another way to do this is to argue that a problem like this must have a constant "doubling time". Since you are told that the initial population doubles in 2 hours, it will double every two hours. In time t, measured in hours, the population will double t/2 times so $P(t)= P(0)2^{t/2}$. (Of course, that is also "cheating" because that is a property of exponential, which is what we get by integrating).

Notice that $t ln(2)= ln(2^t)$ so that [math]e^{t ln(2)}= e^{ln(2^t)}= 2^t[/itex].