1. Population Question

A population, $P$, is said to be growing logistically if the time, $T$, taken for it to increase from $P_1$ and $P_2$ is given by:

$T=\int_{P_1}^{P_2} \frac {kdP}{P(L-P)}$

where $k$ and $L$ are positive constants and $P_1 < P_2 < L$.

A) Calculate the time taken for the population to grow from $P_1 = \frac {L}{4}$ to $P_2 = \frac {L}{2}$.

B)What happens to $T$ as $P_2$ approaches $L$?

I honestly have no clue how to even start this problem. Any help would be appreciated.

Thanks!

2. Originally Posted by tiace
A population, $P$, is said to be growing logistically if the time, $T$, taken for it to increase from $P_1$ and $P_2$ is given by:

$\int_{P_1}^{P_2} \frac {kdP}{P(L-P)}$

where $k$ and $L$ are positive constants and $P_1 < P_2 < L$.

A) Calculate the time taken for the population to grow from $P_1 = \frac {L}{4}$ to $P_2 = \frac {L}{2}$.

B)What happens to $T$ as $P_2$ approaches $L$?

I honestly have no clue how to even start this problem. Any help would be appreciated.

Thanks!
To do the integration you'll need to use Partial Fractions. Then substitute $P_1 = \frac{L}{4}$ and $P_2 = \frac{L}{2}$ as your terminals.

3. Is this integral correct?

$T=\int_{P_1}^{P_2} \frac {kdP}{P(L-P)} = \frac {-k}{L} * (\ln|P-L|-\ln|P|)$

I'm not sure where to go from here though.

4. Yes that's fine.

So now substitute your terminals.

5. I'm not sure what you mean by substitute your terminals.

Do you mean:

$T=(\frac {-k}{L} * (\ln|\frac{L}{2}-L|-\ln|\frac{L}{2}|))-(\frac {-k}{L} * (\ln|\frac{L}{4}-L|-\ln|\frac{L}{4}|))$

Which simplifies to:

$T=\frac{k\ln{3}}{L}$