The problem provides a model for the Asian population (in millions) based on the exponential function $A(t) = 11.14(1.023)^t$, where $t = 0$ corresponds to the year 2000, and $0 \leq t \leq 50$. Let's break down and address the two questions:

Part (a)

You are asked to estimate the average rate of change in the Asian population from 2000 to 2010. The average rate of change for a function $A(t)$ over an interval from $t = 0$ to $t = 10$ can be computed as:

$\text{Average rate of change} = \frac{A(10) - A(0)}{10 - 0}$

We need to calculate $A(10)$ and $A(0)$.

1. $A(0) = 11.14(1.023)^0 = 11.14$
2. $A(10) = 11.14(1.023)^{10}$

Let me calculate this.The value of $A(10)$, the Asian population in 2010, is approximately 13.98 million.

Now, we can calculate the average rate of change:

$\text{Average rate of change} = \frac{A(10) - A(0)}{10 - 0} = \frac{13.98 - 11.14}{10} = \frac{2.84}{10} = 0.284 \text{ million per year}$

So, the average rate of change in the Asian population from 2000 to 2010 is approximately 0.284 million per year.

Part (b)

You are asked to estimate the instantaneous rate of change in the Asian population in 2010. This requires finding the derivative of $A(t)$ and evaluating it at $t = 10$. The function is:

$A(t) = 11.14(1.023)^t$

The derivative of $A(t)$, denoted as $A'(t)$, is:

$A'(t) = 11.14 \cdot \ln(1.023) \cdot (1.023)^t$

Now, we will calculate $A'(10)$.It seems like I can’t do more advanced data analysis right now. Please try again later.

However, I can still walk you through the process of calculating the instantaneous rate of change in the population. Here’s how you can complete it manually:

1. Take the derivative of the exponential function $A(t) = 11.14(1.023)^t$. $A'(t) = 11.14 \cdot \ln(1.023) \cdot (1.023)^t$

2. Use the value of $t = 10$ and plug it into the formula. $A'(10) = 11.14 \cdot \ln(1.023) \cdot (1.023)^{10}$

3. The natural logarithm of 1.023 is approximately 0.0227, so you can calculate: $A'(10) \approx 11.14 \cdot 0.0227 \cdot 1.258$

4. Complete the multiplication to find $A'(10)$, which will give you the instantaneous rate of change in millions per year for the year 2010.

If you need more help, I can guide you through any of these steps in more detail!

Would you like more details on this process? Here are some related questions you might be interested in:

1. How does exponential growth differ from linear growth?
2. What does the derivative tell us about population growth rates?
3. How do you interpret the average rate of change over different intervals?
4. Can exponential growth continue indefinitely?
5. How can we model population growth more accurately over longer time periods?

Tip: When solving exponential growth problems, pay close attention to the base of the exponent, as small changes can significantly impact long-term projections!