Respuesta :
Answer:
Therefore logistic equation is
[tex]P(t) = \frac{4000}{1+ 24(\frac{36}{11})^t}[/tex]
Step-by-step explanation:
The logistic equation is
[tex]\frac{dP}{dt}= kP(1-\frac{P}{k})[/tex]
The analytic solution is
[tex]P(t)= \frac{K}{1+Ae^{-kt}}[/tex] ; [tex]A=\frac{K-P_0}{P_0}[/tex]
P(t) = the population at time t.
K = the carrying capacity = 4,000
k= constant proportionality
t= time
[tex]P_0[/tex] = 160.
The number of fish tripled in the first year.
P(1) = 3×160 =480
[tex]A= \frac{4000-160}{160}[/tex] =24
[tex]\therefore P(t) =\frac{4000}{1+24 e^{-kt}}[/tex] .....(1)
When t= 1 , P(1)=480
[tex]480=\frac{4000}{1+24 e^{-k}}[/tex]
[tex]\Rightarrow 1+24e^{-k} =\frac{4000}{480}[/tex]
[tex]\Rightarrow 24e^{-k} =\frac{4000}{480} -1[/tex]
[tex]\Rightarrow 24e^{-k}= \frac{25}{3}-1[/tex]
[tex]\Rightarrow 24e^{-k}=\frac{22}{3}[/tex]
[tex]\Rightarrow e^{-k} = \frac{22}{3\times 24}[/tex]
[tex]\Rightarrow e^{-k} = \frac{11}{36}[/tex]
taking ln both sides
[tex]\Rightarrow ln e^{-k} =ln \frac{11}{36}[/tex]
[tex]\Rightarrow {-k} = ln\frac{11}{36}[/tex]
[tex]\Rightarrow k = -ln\frac{11}{36}[/tex]
Putting the value of k in equation (1)
[tex]P(t)=\frac{4000}{1+24e^{ln\frac{36}{11}t}}[/tex]
[tex]\Rightarrow P(t) = \frac{4000}{1+24 (e^{ln\frac{36}{11}})^t}[/tex]
[tex]\Rightarrow P(t) = \frac{4000}{1+ 24(\frac{36}{11})^t}[/tex]
Therefore logistic equation is
[tex]P(t) = \frac{4000}{1+ 24(\frac{36}{11})^t}[/tex]
