Question

The future value that accrues when $500 is invested at 7%, compounded continuously, is S(t) = 500e0.07t where t is the number of years. (Round your answers to the nearest cent.) (a) At what rate is the money in this account growing when t = 6? $ per year (b) At what rate is it growing when t = 12? $ per year

Answer 1

(a) $53.27 per year

(b) $81.07 per year

Explanation:

The derivative of the function is ...

S'(t) = 0.07·500e^(0.07t) = 35e^(0.07t)

(a) The derivative evaluated at t=6 is ...

S'(6) = 35·e^0.42 ≈ 53.27

At t=6, the account is growing at the rate of $53.27 per year.

__

(b) The derivative evaluated at t=12 is ...

S'(12) = 35·e^0.84 ≈ 81.07

At t=12, the account is growing at the rate of $81.07 per year.

Answer 2

a) 53.26 $/year

b) 81.07 $/year

Explanation:

Data provided in the question:

Amount invested = $500

Interest rate = 7%

Future value, S(t) =
`500e^(0.07t)`

Now,

rate of growth of money = S'(t) =
`(d(500e^(0.07t)))/(dt)`

or

S'(t) =
`0.07*500e^(0.07t)`

or

S'(t) =
`35e^(0.07t)`

a) at t = 6

S'(t) =
`35e^(0.07(6))`

or

S'(t) =
`35e^(0.42)`

or

S'(t) = 53.26 $/year

b) at t = 12

S'(t) =
`35e^(0.07(12))`

or

S'(t) =
`35e^(0.84)`

or

S'(t) = 81.07 $/year