Question

Suppose that $10,083 is invested at an interest rate of 6.8% 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) The exponential function is
`A(t) = 10083(1.068)^t`

b)

The balance after 1 year is of $10,768.644

The balance after 2 years is of $11,500.91

The balance after 5 years is of $14,010.25.

The balance after 10 years is of $19,467.15

c)

The doubling time is of 10.54 years.

Explanation:

Continuously compounded interest:

The amount of money earning after t years, with interest compounded continuously, is given by:

`A(t) = A(0)(1+r)^t`

In which A(0) is the amount of the initial investment and r is the growth rate, as a decimal.

a) Find the exponential function that describes the amount in the account after time t, in years.

Suppose that $10,083 is invested at an interest rate of 6.8% per year

This means, respectively, that
`A(0) = 10083, r = 0.068`

So

`A(t) = A(0)(1+r)^t`

`A(t) = 10083(1+0.068)^t`

`A(t) = 10083(1.068)^t`

b) What is the balance after 1 year? 2 years? 5 years? 10 years?

After 1 year:

`A(1) = 10083(1.068)^(1) = 10,768.644`

The balance after 1 year is of $10,768.644

After 2 years:

`A(2) = 10083(1.068)^(2) = 11,500.91`

The balance after 2 years is of $11,500.91.

After 5 years:

`A(5) = 10083(1.068)^(5) = 14,010.25`

The balance after 5 years is of $14,010.25.

After 10 years:

`A(10) = 10083(1.068)^(10) = 19,467.15`

The balance after 10 years is of $19,467.15.

c) What is the doubling time?

This is t for which
`A(t) = 2A(0)`. So

`A(t) = A(0)(1.068)^t`

`(1.068)^t = 2`

`\log{(1.068)^t} = \log{2}`

`t\log{1.068} = \log{2}`

`t = \frac{\log{2}}{\log{1.068}}`

`t = 10.54`

The doubling time is of 10.54 years.