Respuesta :
Answer:
[tex]N(T) = \frac{81*40*e^{1.075t}}}{81 + 40*(e^{1.075t} - 1)}[/tex]
Step-by-step explanation:
This problem can be solved by the logistic equation of populations, that is:
[tex]N(t) = \frac{KN_{0}e^{rt}}}{K + N_{0}(e^{rt} - 1)}[/tex]
In which K is the carrying capacity of the population, [tex]N_{0}[/tex] is the initial population, r is the decimal growth rate and t is the period of time.
In this problem, we have that:
When t = 0, the population is 40: This means that [tex]N_{0} = 40[/tex].
The manager of a national park determines that the park can sustain 81 coyotes. This means that [tex]K = 81[/tex].
When t = 1, the population has increased to 60. This means that [tex]N(1) = 60[/tex].
To write the equation, we have to find the value of r.
[tex]N(t) = \frac{KN_{0}e^{rt}}}{K + N_{0}(e^{rt} - 1)}[/tex]
[tex]60 = \frac{81*40*e^{r}}}{81 + 40*(e^{r} - 1)}[/tex]
[tex]60*(41 + 40e^{r}) = 3240e^{r}[/tex]
[tex]2460 + 2400e^{r} = 3240e^{r}[/tex]
[tex]840e^{r} = 2460[/tex]
[tex]e^{r} = 2.93[/tex]
Now we apply ln to both sides, and:
[tex]r = 1.075[/tex]
The answer is:
[tex]N(T) = \frac{81*40*e^{1.075t}}}{81 + 40*(e^{1.075t} - 1)}[/tex]
