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.

1 answer

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.