Compounded continuously?

suppose you want to have $60,000 in 5 years time in a bank account earning 3% interest, compounded continuously:


a) if you make one lump sum deposite now, how much should you deposite?


B) if you deposit money continuously throughout the 5 year period, at what rate should you deposit it?





please show the steps.... thanks|||You didn't say whether the 3% is 'nominal' or 'effective' (annualised). I'll assume it's 'nominal'.





(a) Let the balance = B(t)


where t is time in years.





dB/dt = 0.03 B


so B = A e^(0.03t)


B(0) = A so


B(t) = B(0) e^(0.03t)


and B(0) is the amount to be deposited.





Putting t = 5, we get


B(5) = B(0) e^(0.15)


The amount to be deposited


= B(0)


= B(5) e^(- 0.15)


= $60,000 * e^(- 0.15)


= $51,642.48 (to nearest cent)





(b) dB/dt = C + 0.03B


where C is the rate of deposit (in dollars per year)


B(0) = 0 and


B(5) = 60,000





dB/dt - 0.03B = C


(dB/dt - 0.03B) e^(- 0.03t) = C e^(- 0.03t)


d/dt[B e^(- 0.03t)] = C e^(- 0.03t)


B e^(- 0.03t) = [C/(- 0.03)] e^(- 0.03t) + D


where D is a constant


B = C/(- 0.03) + D e^(0.03t)


= - C/0.03 + D e^(0.03t)





Substituting t = 0 into that equation, we get


0 = - C/0.03 + D


D = C/0.03


Substituting t = 5, we get


60,000 = - C/0.03 + D e^0.15


= (C/0.03) (e^0.15 - 1)


The rate of deposit required = C


= $60,000 * 0.03 / (e^0.15 - 1)


= $11,122.49/year (to the nearest cent)





Note on 'nominal' and 'effective' interest rates:


The effective interest rate is the equivalent rate if interest is compounded only once per year.


The effective interest rate E corresponding to a nominal rate of 3% is given by


(1 + E)^5 = e^(0.15)


1 + E = e^0.03


E = e^0.03 - 1


= approx. 0.030455


= approx. 3.0455%