Respuesta :
Answer:
k=1
P(12)=256
Step-by-step explanation:
The model for population involving birth and death is given as:
[tex]\frac{dP}{dt}=(b-d)P[/tex]
where b=birth rate and d=death rate.
If the birth and death rate are inversely proportional to [tex]\sqrt{P}[/tex]
[tex]b=\frac{A}{\sqrt{P} } \\d=\frac{B}{\sqrt{P} }[/tex] where A and B are constants of variation.
Substituting b and d into our model
[tex]\frac{dP}{dt}=(\frac{A}{\sqrt{P} }-\frac{B}{\sqrt{P} })P\\\frac{dP}{dt}=\frac{k}{\sqrt{P} }P\\[/tex] where A-B=k, another constant
Simplifying using indices
[tex]\frac{dP}{dt}={k}P^{1-\frac{1}{2} }\\\frac{dP}{dt}={k}P^\frac{1}{2} \\[/tex]
Next, we Separate Variables and Integrate both sides
[tex]\frac{dP}{\sqrt{P} }={k}dt\[/tex]
[tex]\int\frac{dP}{\sqrt{P} }=\int{k}dt\[/tex]
[tex]2P^{1/2} =kt+C[/tex] where C is the constant of integration
[tex]P(t) =(\frac{kt}{2} +C)^2[/tex]
When t=0, P(t)=[tex]P_0[/tex], C=[tex]\sqrt{P_0}[/tex]
[tex]P(t) =(\frac{kt}{2} +\sqrt{P_0})^2[/tex] as required.
(b)If [tex]P_0[/tex]=100, t=6 months, P(t)=169
[tex]P(t) =(\frac{kt}{2} +\sqrt{P_0})^2[/tex]
[tex]169 =(\frac{6k}{2} +\sqrt{100})^2\\\sqrt{169} =3k+100\\13=3k+10\\3k=13-10=3\\k=1[/tex]
Since we have found the constant k, we can then calculate the population after 1 year. Note that we use 12 months since we used month earlier to get k.
[tex]P(t) =(\frac{kt}{2} +\sqrt{P_0})^2[/tex]
[tex]P(12) =(\frac{1X12}{2} +\sqrt{100})^2\\=(6+10)^2=256[/tex]
Therefore the population after a year is 256.
