Let's go through both parts of the problem systematically.

Part (a) – Finding an exponential model $n(t) = n_0 2^{t/a}$

• The initial population $n_0$ in 2012 is given as $220,000$.
• The doubling time is 28 years, so $a = 28$.

We can now plug in these values to get the exponential model:

$n(t) = 220,000 \cdot 2^{t/28}$

So, the exponential model is:

$n(t) = 220,000 \cdot 2^{t/28}$

Part (b) – Finding an exponential model $n(t) = n_0 e^{rt}$

In this model, we need to find the growth rate $r$. Since the population doubles every 28 years, we can use the doubling time formula to find $r$.

The formula for the doubling time is:

$2 = e^{r \cdot 28}$

Taking the natural logarithm (ln) of both sides:

$\ln(2) = r \cdot 28$

Solving for $r$:

$r = \frac{\ln(2)}{28}$

Now, let's calculate the value of $r$ and round it to four decimal places.

$r \approx \frac{0.6931}{28} \approx 0.0248$

Now, we can write the exponential model:

$n(t) = 220,000 \cdot e^{0.0248t}$

Final Results:

• (a) Exponential model: $n(t) = 220,000 \cdot 2^{t/28}$
• (b) Exponential model: $n(t) = 220,000 \cdot e^{0.0248t}$

Would you like more details or have any questions?

Related Questions:

1. How can we use the model to predict the bat population in 2040?
2. What is the bat population expected to be in 2025?
3. How long will it take for the bat population to reach 1 million?
4. What is the percentage growth rate of the bat population?
5. How does changing the doubling time affect the population model?

Tip:

For exponential growth, the doubling time is directly related to the growth rate, and understanding the relationship between them helps in solving population growth problems easily.