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Population model: dP/dt = k sqrtP

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1. Homework Statement

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

[tex] \frac{dP}{dt} = k \sqrt{p} [/tex]

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


3. The Attempt at a Solution

[tex] \frac{dP}{\sqrt{p}} = k dt [/tex]


integrate


[tex] 2 \sqrt{P} = k t + A [/tex]


Sub 0 for t C for X


[tex]A=2 \sqrt{C}[/tex]

sub back in

[tex] 2 \sqrt{P} = 2t+4 \sqrt{c} [/tex]

[tex] P = (2 t k+4 \sqrt{C})^2 [/tex]
 
Last edited:

Answers and Replies

743
1
[tex] 2 \sqrt{P} = 2t+4 \sqrt{c} [/tex]
[tex] P = (2 t k+4 \sqrt{C})^2 [/tex]

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 [tex] \sqrt{P}[/tex] 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

[tex] P(t) = \displaystyle\left( \frac{1}{2} kt + \sqrt(C) \right) ^2 [/tex]
 
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I ended up solving it correctly

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

the answer is

[tex] \frac{(kt+2\sqrt{C})^2}{4} [/tex]
 
HallsofIvy
Science Advisor
Homework Helper
41,738
899
1. Homework Statement

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

[tex] \frac{dP}{dt} = k \sqrt{p} [/tex]

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


3. The Attempt at a Solution

[tex] \frac{dP}{\sqrt{p}} = k dt [/tex]


integrate


[tex] 2 \sqrt{P} = k t + A [/tex]


Sub 0 for t C for X


[tex]A=2 \sqrt{C}[/tex]
You were fine up to here.

sub back in

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

[tex] P = (2 t k+4 \sqrt{C})^2 [/tex]
 

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