# Population model: dP/dt = k sqrtP

1. Jan 26, 2007

### Tom McCurdy

1. The problem statement, all variables and given/known data

Suppose that when a certain lake is stocked with a population P of fish, the birth and death rates $$\alpha$$ and $$\beta$$ are inversely proportional to $$\sqrt{P}$$ , so that

$$\frac{dP}{dt} = k \sqrt{p}$$

Find P(t) if P(0)=C

3. The attempt at a solution

$$\frac{dP}{\sqrt{p}} = k dt$$

integrate

$$2 \sqrt{P} = k t + A$$

Sub 0 for t C for X

$$A=2 \sqrt{C}$$

sub back in

$$2 \sqrt{P} = 2t+4 \sqrt{c}$$

$$P = (2 t k+4 \sqrt{C})^2$$

Last edited: Jan 26, 2007
2. Jan 27, 2007

### Kreizhn

$$2 \sqrt{P} = 2t+4 \sqrt{c}$$
$$P = (2 t k+4 \sqrt{C})^2$$

There are a few small problems with this, mostly just technical. Such as the second to last line is missing a k, and the we substitute C for P; there is no X.

However, not so trivial is where did the coefficient of 2 for $$\sqrt{P}$$ go? How did the 2 in 2kt show up randomly? Your mistakes are algebraic and occur after your line "sub back in"

$$P(t) = \displaystyle\left( \frac{1}{2} kt + \sqrt(C) \right) ^2$$

3. Jan 27, 2007

### Tom McCurdy

I ended up solving it correctly

I ended up solving it correctly (2 min before it was due)

$$\frac{(kt+2\sqrt{C})^2}{4}$$

4. Jan 27, 2007

### HallsofIvy

You were fine up to here.

No, you just said $2\sqrt{P}= kt+ A$. If, now, $A= 2\sqrt{C}$, then $2\sqrt{P}= kt+ 2\sqrt{C}$.
Square both sides: $4 P= (kt+ 2\sqrt{C})^2$
$$P= \frac{kt+ 2\sqrt{C})^2}{4}$$
which is what you say you eventually got