- #1
Mirole
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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 separate 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?
(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)
---------------------------------------------------------
For (a), I could just separate 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?