Question

Compound interest The balance A (in dollars) in a saving account is given by A = 5000e0.08t,Where t is measured in years.Find the rates at which the balance is changing when (a) t = 1 year,and (c) t = 50 years.

Answer 1

a)
`(dA)/(dt)=400 e^(0.08(1)) =433.315`

b)
`(dA)/(dt)=400 e^(0.08(10)) =890.216`

c)
`(dA)/(dt)=400 e^(0.08(50)) =21839.260`

Explanation:

For this case we have the following expression for the balance:

`A = 5000 e^(0.08t)`

And we are interested on the rate at which the balance is changing for some values of t. The correct way to answer this is finding the derivate respect to t of A like this:

`(dA)/(dt)= (d)/(dt) (5000 e^(0.08t))`

`(dA)/(dt)= 5000 (d)/(dt)(e^(0.08t))`

`(dA)/(dt)= 5000*0.08 e^(0.08t)`

`(dA)/(dt)=400 e^(0.08t)`

So then we can solve for the different years like this

Part a

t= 1 years

`(dA)/(dt)=400 e^(0.08(1)) =433.315`

So on this case would be $433.315 dollars per year

Part b

t=10 years

`(dA)/(dt)=400 e^(0.08(10)) =890.216`

So on this case would be $890.216 dollars per year

Part c

t=50 years

`(dA)/(dt)=400 e^(0.08(50)) =21839.260`

So on this case would be $21839.260 dollars per year