Population model: dP/dt = k sqrtP

[Tom McCurdy]

Homework Statement

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

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

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

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

 

[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"

Thus your solutions should be

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

 

[Tom McCurdy]

I ended up solving it correctly

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

the answer is

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

 

[HallsofIvy]

Homework Statement

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

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


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

You were fine up to here.

sub back in

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

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

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