Question

Suppose that $19,911 is invested at an interest rate of 6.1% per year, compounded continuously. a) Find the exponential function that describes the amount in the account after time t, in years. b) What is the balance after 1 year? 2 years? 5 years? 10 years? c) What is the doubling time?

Answer 1

a) a(t) = 19911e^(0.061t)

b) 1 yr: $21163.38; 2 yr: $22494.53; 5 yr: $27011.76; 10 yr: $36644.83

c) 11.36 years

Explanation:

You want the exponential function that describes account value of a $19911 investment earning 6.1% interest compounded continuously, and its value for 1, 2, 5, and 10 years. You also want the account's doubling time.

a) Function

The amount a(t) in the account will be given by the exponential function ...

a(t) = (initial amount)·e^(rt) . . . . . for annual interest rate r and t years

a(t) = 19911·e^(0.061t)

b) Balance

The attached calculator display shows the balances to be ...

• 1 yr: $21163.38
• 2 yr: $22494.53
• 5 yr: $27011.76
• 10 yr: $36644.83

c) Doubling time

For continuous compounding, the doubling time is given by ...

t = 69.315/r . . . . . where r is in percent

t = 69.315/6.1 ≈ 11.363

It will take about 11.36 years for the balance to double.

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