Math Problem Statement

When a breeding group of animals is introduced into a restricted area such as a wildlife reserve, the population can be expected to grow rapidly at first, but to level out when the population grows to near the maximum that the environment can support. Such growth is known as logistic population growth, and ecologists sometimes use a formula to describe it. The number N of deer present at time t (measured in years since the herd was introduced) on a certain wildlife reserve has been determined by ecologists to be given by the function below. (Round your answers down to the nearest whole number.) N = 12.31 0.03 + 0.58t

(a) How many deer were initially on the reserve?

(b) Calculate N(12). N(12) =

(c) Express the number of deer present after 17 years using functional notation. N(17)=

Calculate the number of deer present after 17 years.

(d) How much increase in the deer population do you expect from the 12th to the 17th year?

Solution

The function provided describes the population N(t)N(t) of deer on a wildlife reserve as a function of time tt (in years). The function is:

N(t)=12.310.03+0.58tN(t) = \frac{12.31}{0.03 + 0.58t}

Let's address each part of the question.

(a) Initial Population

The initial population of deer corresponds to t=0t = 0.

N(0)=12.310.03+0.58(0)=12.310.03N(0) = \frac{12.31}{0.03 + 0.58(0)} = \frac{12.31}{0.03}

Let's calculate:

N(0)=12.310.03410.33N(0) = \frac{12.31}{0.03} \approx 410.33

Rounding down to the nearest whole number:

N(0)410 deerN(0) \approx 410 \text{ deer}

(b) Calculate N(12)N(12)

We need to find the population after 12 years, which is N(12)N(12).

N(12)=12.310.03+0.58(12)=12.310.03+6.96=12.316.99N(12) = \frac{12.31}{0.03 + 0.58(12)} = \frac{12.31}{0.03 + 6.96} = \frac{12.31}{6.99}

Let's calculate:

N(12)12.316.991.76N(12) \approx \frac{12.31}{6.99} \approx 1.76

Rounding down to the nearest whole number:

N(12)1 deerN(12) \approx 1 \text{ deer}

(c) Express the number of deer present after 17 years using functional notation

The number of deer after 17 years is N(17)N(17). We calculate:

N(17)=12.310.03+0.58(17)=12.310.03+9.86=12.319.89N(17) = \frac{12.31}{0.03 + 0.58(17)} = \frac{12.31}{0.03 + 9.86} = \frac{12.31}{9.89}

Let's calculate:

N(17)12.319.891.24N(17) \approx \frac{12.31}{9.89} \approx 1.24

Rounding down to the nearest whole number:

N(17)1 deerN(17) \approx 1 \text{ deer}

(d) Population Increase from Year 12 to Year 17

The increase in population from year 12 to year 17 is the difference N(17)N(12)N(17) - N(12).

N(17)N(12)=11=0 deerN(17) - N(12) = 1 - 1 = 0 \text{ deer}

So, there is no expected increase in the deer population from the 12th to the 17th year.


Would you like further details or have any questions?

Here are 5 related questions you could explore:

  1. How does the logistic growth model differ from exponential growth?
  2. What factors could affect the carrying capacity in this scenario?
  3. How would the population change if the carrying capacity were altered?
  4. How can you determine the inflection point of the logistic growth curve?
  5. What are real-world examples of logistic population growth?

Tip: When dealing with population models, always consider the environmental factors that might limit growth, as they play a crucial role in shaping the population curve.

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Math Problem Analysis

Mathematical Concepts

Logistic Growth
Rational Functions

Formulas

N(t) = 12.31 / (0.03 + 0.58t)

Theorems

Logistic Growth Model

Suitable Grade Level

Grades 11-12