Question

The function given by y= f(x) shows the value of $5000 invested at 6% Interest compounded continuously, x years after the money was originally invested.
(Round your answers to the nearest cent.)
Value of $5000 with Continuous Compounding at 6%
(a) Find the average amount earned per year between the 5th year and the 10th year.

Answer 1

To find the average amount earned per year between the 5th year and the 10th year for an investment of $5000 compounded continuously at 6%, we first need to use the formula for continuous compounding:

\[
A(t) = Pe^{rt}
\]

Where:
- \( A(t) \) = amount of money accumulated after time \( t \) (in years)
- \( P \) = principal amount (initial investment)
- \( r \) = annual interest rate (as a decimal)
- \( t \) = time in years
- \( e \) = Euler's number (approximately 2.71828)

Given:
- \( P = 5000 \)
- \( r = 0.06 \)

1. **Calculate the amount after 5 years**:
\[
A(5) = 5000 e^{0.06 \cdot 5}
\]
\[
A(5) = 5000 e^{0.3}
\]
Using a calculator to find \( e^{0.3} \):
\[
e^{0.3} \approx 1.34986
\]
So,
\[
A(5) = 5000 \cdot 1.34986 \approx 6749.30
\]

2. **Calculate the amount after 10 years**:
\[
A(10) = 5000 e^{0.06 \cdot 10}
\]
\[
A(10) = 5000 e^{0.6}
\]
Using a calculator to find \( e^{0.6} \):
\[
e^{0.6} \approx 1.82212
\]
So,
\[
A(10) = 5000 \cdot 1.82212 \approx 9110.61
\]

3. **Determine the total amount earned between the 5th year and the 10th year**:
The total earned between these two years is:
\[
\text{Total earned} = A(10) - A(5)
\]
\[
\text{Total earned} = 9110.61 - 6749.30 \approx 2361.31
\]

4. **Calculate the average amount earned per year between the 5th year and the 10th year**:
The average amount earned per year over the 5 years (from year 5 to year 10) is:
\[
\text{Average per year} = \frac{\text{Total earned}}{\text{Number of years}} = \frac{2361.31}{5} \approx 472.26
\]

Thus, the average amount earned per year between the 5th year and the 10th year is approximately **$472.26**.