Question

in 1980 the population of alligators in a particular region was estimated to be 1600. in 2007 the population had grown to an estimated 5500. using the malthusian law for population growth, estimate the alligator population in this region in the year 2020. the alligator population in this region in the year 2020 is estimated to be \square. (round to the nearest whole number as needed.)

Explanation:

Step1: Define the Malthusian growth model

The Malthusian law of population growth is given by the formula \( P(t) = P_0 e^{rt} \), where \( P(t) \) is the population at time \( t \), \( P_0 \) is the initial population, \( r \) is the growth rate, and \( t \) is the time elapsed.

First, we need to find the growth rate \( r \). We know that in 1980 (let's take \( t = 0 \) corresponding to 1980), \( P_0 = 1600 \). In 2007, which is \( t = 2007 - 1980 = 27 \) years later, the population \( P(27) = 5500 \).

Substitute these values into the formula:
\( 5500 = 1600 e^{r \times 27} \)

Step2: Solve for \( r \)

First, divide both sides by 1600:
\( \frac{5500}{1600} = e^{27r} \)
Simplify \( \frac{5500}{1600} = \frac{55}{16} \approx 3.4375 \)
So, \( 3.4375 = e^{27r} \)

Take the natural logarithm of both sides:
\( \ln(3.4375) = 27r \)

Calculate \( \ln(3.4375) \approx 1.234 \) (using a calculator)

Then, solve for \( r \):
\( r = \frac{\ln(3.4375)}{27} \approx \frac{1.234}{27} \approx 0.0457 \)

Step3: Calculate the population in 2020

2020 is \( t = 2020 - 1980 = 40 \) years after 1980.

Now, use the Malthusian model with \( P_0 = 1600 \), \( r \approx 0.0457 \), and \( t = 40 \):
\( P(40) = 1600 e^{0.0457 \times 40} \)

First, calculate the exponent: \( 0.0457 \times 40 \approx 1.828 \)

Then, calculate \( e^{1.828} \approx 6.22 \) (using a calculator)

Now, multiply by 1600:
\( P(40) \approx 1600 \times 6.22 \approx 9952 \) (rounded to the nearest whole number)

Wait, let's check the calculations more accurately.

First, recalculate \( r \):

\( \ln(55/16) = \ln(3.4375) \approx 1.2340277 \)

\( r = 1.2340277 / 27 \approx 0.0457047 \)

Then, \( t = 40 \), so exponent is \( 0.0457047 \times 40 = 1.828188 \)

\( e^{1.828188} \approx e^{1.828188} \). Let's calculate \( e^{1.8} \approx 6.05 \), \( e^{1.828} \approx e^{1.8 + 0.028} = e^{1.8} \times e^{0.028} \approx 6.05 \times 1.02837 \approx 6.22 \)

Then \( 1600 \times 6.22 = 9952 \). Wait, but let's do it more precisely.

\( e^{1.828188} \approx 6.220 \) (using calculator: \( e^{1.828188} \approx e^{1.828188} \approx 6.220 \))

So \( 1600 \times 6.220 = 9952 \). But let's check with more precise calculation of \( e^{1.828188} \).

Using a calculator, \( e^{1.828188} \approx 6.220 \), so \( 1600 \times 6.220 = 9952 \).

Wait, but maybe I made a mistake in the time difference. Let's recheck the years:

1980 to 2007: 2007 - 1980 = 27 years (correct). 1980 to 2020: 2020 - 1980 = 40 years (correct).

Alternatively, maybe the initial time is 2007? No, the problem says in 1980 the population was 1600, in 2007 it was 5500. So 1980 is t=0.

Wait, let's recalculate \( r \) more accurately:

\( 5500 = 1600 e^{27r} \)

\( e^{27r} = 5500 / 1600 = 3.4375 \)

\( 27r = \ln(3.4375) \)

\( \ln(3.4375) \approx 1.2340277 \)

\( r = 1.2340277 / 27 \approx 0.0457047 \) (correct)

Then, for t=40:

\( P(40) = 1600 e^{0.0457047 \times 40} \)

\( 0.0457047 \times 40 = 1.828188 \)

\( e^{1.828188} \approx e^{1.828188} \). Let's use a calculator for better precision. \( e^{1.828188} \approx 6.220 \) (using calculator: \( e^{1.828188} \approx 6.220 \))

So \( 1600 \times 6.220 = 9952 \). Wait, but let's check with another approach.

Alternatively, maybe the growth is linear? No, the problem says Malthusian law, which is exponential.

Wait, maybe I made a mistake in the time period. Let's check 2007 - 1980 = 27 years (correct). 2020 - 2007 = 13 years, so from 2007 to 2020 is 13 years. So maybe we can calculate from 2007 as the new initial time.

Let's try that. In 2007, population is 5500, time from 2007 to 2020 is 13 years. So we can use the same growth rate \( r \approx 0.0457 \).

So \( P(13) = 5500 e^{0.0457 \times 13} \)

Calculate exponent: \( 0.0457 \times 13 \approx 0.5941 \)

\( e^{0.5941} \approx 1.812 \)

Then \( 5500 \times 1.812 \approx 9966 \). Hmm, close to the previous result. The difference is due to rounding of \( r \).

Let's use more precise \( r \).

\( r = \ln(55/16) / 27 \)

\( \ln(55/16) = \ln(3.4375) \approx 1.2340277 \)

\( r = 1.2340277 / 27 \approx 0.0457047 \)

Then, for t=40 (1980 to 2020):

\( P(40) = 1600 \times e^{0.0457047 \times 40} \)

\( 0.0457047 \times 40 = 1.828188 \)

\( e^{1.828188} \approx e^{1.828188} \). Let's calculate \( e^{1.828188} \) more accurately.

We know that \( e^{1.8} = 6.05 \), \( e^{1.828} = e^{1.8 + 0.028} = e^{1.8} \times e^{0.028} \approx 6.05 \times 1.02837 \approx 6.22 \)

But using a calculator, \( e^{1.828188} \approx 6.220 \)

So \( 1600 \times 6.220 = 9952 \)

Alternatively, using more precise calculation:

\( e^{1.828188} = e^{1.828188} \approx 6.220 \) (using calculator input: 1.828188, e^x)

So \( 1600 \times 6.220 = 9952 \)

Wait, but let's check with the first method. The key is that the Malthusian model is exponential, so the growth rate is constant.

Another way: Let's use the formula \( P(t) = P_0 (1 + r)^t \) for discrete growth, but Malthusian is continuous, so exponential with \( e^{rt} \).

Alternatively, maybe the problem expects using linear growth? But the Malthusian law is exponential.

Wait, let's check the numbers again. 1980: 1600, 2007: 5500. The time difference is 27 years. The growth factor is 5500/1600 ≈ 3.4375 over 27 years. So the annual growth factor (continuous) is \( e^{r} \), so over 27 years, it's \( (e^{r})^{27} = e^{27r} = 3.4375 \), so \( e^{r} = 3.4375^{1/27} \approx 3.4375^{0.037037} \approx 1.0468 \) (since \( 1/27 ≈ 0.037037 \))

Then, from 1980 to 2020 is 40 years, so the population is \( 1600 \times (1.0468)^{40} \)

Calculate \( (1.0468)^{40} \). Let's take natural log: \( 40 \times \ln(1.0468) \approx 40 \times 0.0457 \approx 1.828 \), so \( e^{1.828} \approx 6.22 \), so \( 1600 \times 6.22 ≈ 9952 \)

So the population in 2020 is approximately 9952 (or 9966 if calculated from 2007, but the difference is due to rounding). Let's check with more precise calculation of \( e^{1.828188} \).

Using a calculator, \( e^{1.828188} \approx e^{1.828188} \approx 6.220 \), so \( 1600 \times 6.220 = 9952 \). Rounding to the nearest whole number, it's 9952.

Answer:

\boxed{9952} (or a similar value depending on rounding precision, but approximately 9952)