Respuesta :
Answer:
The equation to find the number of fish after t years where y is the number of fish is:
y(t) = 160×3^(t) ( t <= 2.93 assuming that the maximum number of fish is 4000 )
Therefore when y(t) = 2000 2000 = 160×3^(t)
3^(t) = 25/2 log 3 (3^(t)) = log 3 (25/2) t = 2.29 years.
Using the logistic equation, we have that:
a)
The equation is:
[tex]P(t) = \frac{4000}{1 + 24e^{-1.1856t}}[/tex]
b)
It will take 2.68 years for the population to increase to 2000.
The logistic equation is:
[tex]P(t) = \frac{K}{1 + Ae^{-kt}}[/tex]
With:
[tex]A = \frac{K - P(0)}{P(0)}[/tex]
The parameters are:
- The carrying capacity K.
- The decay rate k.
- The initial population P(0).
In this problem:
- Initial population of 160, thus [tex]P(0) = 160[/tex].
- Carrying capacity of 4,000, thus [tex]K = 4000[/tex].
Then:
[tex]A = \frac{4000 - 160}{160} = 24[/tex]
Thus:
[tex]P(t) = \frac{4000}{1 + 24e^{-kt}}[/tex]
Item a:
Tripled during the first year, thus [tex]P(1) = 3P(0) = 3(160) = 480[/tex].
This is used to find k.
[tex]480 = \frac{4000}{1 + 24e^{-k}}[/tex]
[tex]480 + 11520e^{-k} = 4000[/tex]
[tex]e^{-k} = \frac{3520}{11520}[/tex]
[tex]\ln{e^{-k}} = \ln{\frac{3520}{11520}}[/tex]
[tex]k = -\ln{\frac{3520}{11520}}[/tex]
[tex]k = 1.1856[/tex]
Thus, the equation is:
[tex]P(t) = \frac{4000}{1 + 24e^{-1.1856t}}[/tex]
Item b:
This is t for which P(t) = 2000, thus:
[tex]P(t) = \frac{4000}{1 + 24e^{-1.1856t}}[/tex]
[tex]2000 = \frac{4000}{1 + 24e^{-1.1856t}}[/tex]
[tex]\frac{1}{1 + 24e^{-1.1856t}} = 0.5[/tex]
[tex]0.5 + 12e^{-1.1856t} = 1[/tex]
[tex]e^{-1.1856t} = \frac{1}{24}[/tex]
[tex]\ln{e^{-1.1856t}} = \ln{\frac{1}{24}}[/tex]
[tex]t = -\frac{\ln{\frac{1}{24}}}{1.1856t}[/tex]
[tex]t = 2.68[/tex]
It will take 2.68 years for the population to increase to 2000.
A similar problem is given at https://brainly.com/question/24215157
