1.

A person places Rs 500 in an account that interest compounded continuously. Assuming no additional deposits or withdrawals, how much will be in the account after seven years if the interest rate isa constant 8.5 percent for the first four year and a constant 9.25 percent for the last three years

Answer» Let N(t) denote the balance in the account at any time t.
Initially, N(0)=500.
For the first four years, k=0.085 N=0
Its solution is `N(t)=ce^(0.085t) (0 letle4)`………….(1)
At t-0, N(0)=500. Then from (1), `5000=ce^(0.085(0))=c`
and equation (1) becomes `N(t) = 5000e^(0.085t) (0 letle4)`...................(2)
Substituting t=4 into equation (2), we find the balance after four years to be
`N(4) = 5000e^(0.085(4))=5000(1.404948)=7024.74`.
This amount also represents the beginning balance for the last three-year period.
Over the last three years, the interest rate is `9.25%`
`therefore (dN)(dt) -0.0925 N=0` `(4 let le7)`
In solution is `N(t)=ce^(0.0925t) (4 letle7)`..........(3)
At `t=4, N(4)=7024.74`. Then from equations (3),
`7024.74=ce^(0.009245) =c(1.447735)` or `c=4852.23`
Then from equation (3),
`N(7) = 4852.23 e^(0.0925(7))=4852.23(1.910758)`


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