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§ 5.2

§ 5.2. Compound Interest. Section Outline. Compound Interest: Non-Continuous Compound Interest: Continuous Applications of Interest Compounded Continuously. Compound Interest: Non-Continuous. Compound Interest: Non-Continuous. P = principal amount invested

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§ 5.2

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  1. §5.2 Compound Interest

  2. Section Outline • Compound Interest: Non-Continuous • Compound Interest: Continuous • Applications of Interest Compounded Continuously

  3. Compound Interest: Non-Continuous Compound Interest: Non-Continuous • P = principal amount invested • m = the number of times per year interest is compounded • r = the interest rate • t = the number of years interest is being compounded • A = the compound amount, the balance after t years

  4. Compound Interest Notice that as m increases, so does A. Therefore, the maximum amount of interest can be acquired when m is being compounded all the time - continuously.

  5. Compound Interest: Continuous Compound Interest: Continuous • P = principal amount invested • r = the interest rate • t = the number of years interest is being compounded • A = the compound amount, the balance after t years

  6. Compound Interest: Continuous EXAMPLE (Continuous Compound) Ten thousand dollars is invested at 6.5% interest compounded continuously. When will the investment be worth $41,787? SOLUTION We must first determine the formula for A(t). Since interest is being compounded continuously, the basic formula to be used is Since the interest rate is 6.5%, r = 0.065. Since ten thousand dollars is being invested, P = 10,000. And since the investment is to grow to become $41,787, A = 41,787. We will make the appropriate substitutions and then solve for t. This is the formula to use. P = 10,000, r = 0.065, and A = 41,787.

  7. Compound Interest: Continuous CONTINUED Divide by 10,000. Rewrite the equation in logarithmic form. Divide by 0.065 and solve for t. Therefore, the $10,000 investment will grow to $41,787, via 6.5% interest compounded continuously, in 22 years.

  8. Compound Interest: Present Value Variables are defined the same as in Slide #21.

  9. Compound Interest: Present Value EXAMPLE (Investment Analysis) An investment earns 5.1% interest compounded continuously and is currently growing at the rate of $765 per year. What is the current value of the investment? SOLUTION Since the problem involves a rate of change, we will use the formula for the derivative of That is, A΄ = rA. Since the investment is growing at a rate of $765 per year, A΄ = 765. Since the interest rate is 5.1%, r = 0.051. This is the given function. A΄ = 765 and r = 0.051. Solve for A. Therefore, the value of A for this situation is 15,000. We can now use this, and the present value formula, to determine P.

  10. Compound Interest: Present Value CONTINUED This is the present value formula. A = 15,000, r = 0.051 and t = 1 (since we were given the rate of growth per year). Simplify. Therefore, the current value is $14,254.18.

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