Question

The manager of a national park determines that the park can sustain 81 coyotes. The manager assumes that the rate of change of the number of coyotes N is directly proportional to the difference between the maximum population and the current population. When t = 0, the population is 40, and t = 1, the population has increased to 60. Write the population function N as a function of t.

Answer 1

`N(T) = \frac{81*40*e^(1.075t)}}{81 + 40*(e^(1.075t) - 1)}`

Explanation:

This problem can be solved by the logistic equation of populations, that is:

`N(t) = \frac{KN_(0)e^(rt)}}{K + N_(0)(e^(rt) - 1)}`

In which K is the carrying capacity of the population,
`N_(0)` 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
`N_(0) = 40`.

The manager of a national park determines that the park can sustain 81 coyotes. This means that
`K = 81`.

When t = 1, the population has increased to 60. This means that
`N(1) = 60`.

To write the equation, we have to find the value of r.

`N(t) = \frac{KN_(0)e^(rt)}}{K + N_(0)(e^(rt) - 1)}`

`60 = \frac{81*40*e^(r)}}{81 + 40*(e^(r) - 1)}`

`60*(41 + 40e^(r)) = 3240e^(r)`

`2460 + 2400e^(r) = 3240e^(r)`

`840e^(r) = 2460`

`e^(r) = 2.93`

Now we apply ln to both sides, and:

`r = 1.075`

The answer is:

`N(T) = \frac{81*40*e^(1.075t)}}{81 + 40*(e^(1.075t) - 1)}`