# Thread: Population Question

1. ## Population Question

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

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

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

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

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

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

Thanks!

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

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

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

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

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

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 $\displaystyle P_1 = \frac{L}{4}$ and $\displaystyle P_2 = \frac{L}{2}$ as your terminals.

3. Is this integral correct?

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

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:

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

Which simplifies to:

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