To solve this problem, let's break it down step by step.

Part 1: Formula for Average Rate of Change

The average rate of change of a function $A(t)$ from $t = a$ to $t = b$ is given by the formula:

$\text{Average Rate of Change} = \frac{A(b) - A(a)}{b - a}$

For the interval from $t = 0$ to $t = 5$, this becomes:

$\text{Average Rate of Change} = \frac{A(5) - A(0)}{5}$

This matches with Option B in your question, which is correct.

Part 2: Calculating the Average Rate of Change for the First 5 Years

The total amount in the account after $t$ years is given by the formula:

$A(t) = 1100 \times (1.05)^t$

Step 1: Calculate $A(0)$ and $A(5)$

• $A(0)$: At $t = 0$, the amount is simply the initial investment. $A(0) = 1100 \times (1.05)^0 = 1100$

• $A(5)$: At $t = 5$, we calculate the amount as: $A(5) = 1100 \times (1.05)^5$

Let's calculate $A(5)$.

$A(5) \approx 1100 \times 1.27628 = 1403.91$

Step 2: Find the Average Rate of Change

Now, use the formula:

$\text{Average Rate of Change} = \frac{A(5) - A(0)}{5} = \frac{1403.91 - 1100}{5} = \frac{303.91}{5} = 60.782$

So, the average rate of change for the first 5 years is approximately $60.78 per year.

b. Average Rate of Change for the Second 5 Years (from $t = 5$ to $t = 10$)

Step 1: Calculate $A(10)$

$A(10) = 1100 \times (1.05)^{10}$

Let's calculate $A(10)$.

$A(10) \approx 1100 \times 1.62889 = 1791.78$

Step 2: Calculate the Average Rate of Change for $t = 5$ to $t = 10$

$\text{Average Rate of Change} = \frac{A(10) - A(5)}{5} = \frac{1791.78 - 1403.91}{5} = \frac{387.87}{5} = 77.574$

So, the average rate of change for the second 5 years is approximately $77.57 per year.

c. Estimate the Instantaneous Rate of Change at $t = 5$

The instantaneous rate of change is the derivative of $A(t)$ at $t = 5$.

Step 1: Find the Derivative $A'(t)$

$A'(t) = 1100 \times \frac{d}{dt}(1.05)^t = 1100 \times (1.05)^t \times \ln(1.05)$

Step 2: Calculate $A'(5)$

$A'(5) \approx 1100 \times (1.05)^5 \times \ln(1.05) \approx 1403.91 \times 0.04879 = 68.482$

So, the instantaneous rate of change at $t = 5$ is approximately $68.48 per year.

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Would you like more details or have any further questions?

Here are some related questions you might find interesting:

1. How does the compounding frequency affect the average rate of change?
2. What is the total amount after 20 years with the same interest rate?
3. How would the average rate of change differ if the interest were compounded monthly?
4. Can we estimate the instantaneous rate of change using a different method, like the secant line approach?
5. How does the logarithmic function in the derivative formula contribute to the instantaneous rate?

Tip: The average rate of change gives a broad view over an interval, while the instantaneous rate of change is like a snapshot at a specific point in time, useful for understanding how quickly the investment grows at that exact moment.