To solve the problem, we'll start with part (a) and then move on to part (b).
(a) Function that models the population
The population growth can be modeled by the formula for continuous exponential growth:
\[ P(t) = P_0 \cdot e^{rt} \]
where:
- \( P_0 \) is the initial population,
- \( r \) is the growth rate (as a decimal),
- \( t \) is the time in years since the initial time period.
In this case:
- \( P_0 = 20800 \) (the population in the year 2000),
- \( r = 0.06 \) (6% expressed as a decimal),
- \( t \) is the number of years after 2000.
Thus, the function becomes:
\[ P(t) = 20800 \cdot e^{0.06t} \]
(b) Estimate the fox population in the year 2008
To find the population in the year 2008, we need to calculate \( P(t) \) where \( t = 8 \) (since 2008 is 8 years after 2000):
\[ P(8) = 20800 \cdot e^{0.06 \cdot 8} \]
Calculating the exponent:
\[ 0.06 \cdot 8 = 0.48 \]
Now we compute \( P(8) \):
\[ P(8) = 20800 \cdot e^{0.48} \]
Using a calculator to find \( e^{0.48} \):
\[ e^{0.48} \approx 1.615 \]
Now substituting back into the population equation:
\[ P(8) = 20800 \cdot 1.615 \approx 33632 \]
Since we need to round to the nearest integer, the estimated fox population in 2008 is approximately 33632.
Thus, the final answers are:
- (a) \( P(t) = 20800 \cdot e^{0.06t} \)
- (b) The estimated fox population in 2008 is 33632.