Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 1 of 47 § 5.2 Compound Interest.

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Presentation transcript:

Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 1 of 47 § 5.2 Compound Interest

Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 2 of 47  Compound Interest: Non-Continuous  Compound Interest: Continuous  Applications of Interest Compounded Continuously Section Outline

Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 3 of 47 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

Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 4 of 47 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.

Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 5 of 47 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

Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 6 of 47 Compound Interest: ContinuousEXAMPLE SOLUTION (Continuous Compound) Ten thousand dollars is invested at 6.5% interest compounded continuously. When will the investment be worth $41,787? 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 = 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.

Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 7 of 47 Compound Interest: Continuous Therefore, the $10,000 investment will grow to $41,787, via 6.5% interest compounded continuously, in 22 years. Divide by 10,000. CONTINUED Rewrite the equation in logarithmic form. Divide by and solve for t.

Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 8 of 47 Compound Interest: Present Value Variables are defined the same as in Slide #21.

Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 9 of 47 Compound Interest: Present ValueEXAMPLE SOLUTION (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? 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 = This is the given function. A΄ = 765 and r = 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.

Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 10 of 47 Compound Interest: Present Value This is the present value formula. Therefore, the current value is $14, CONTINUED A = 15,000, r = and t = 1 (since we were given the rate of growth per year). Simplify.