Question

1. Suppose you have A, dollars to invest in a savings account eaming an annual interest rate of r percent compounded continuously. Furthermore, suppose that you make annual deposits of d dollars to the account. The differential equation governing this situation is dA =rA+d, AO) = Ao (a) Find an equation for the future value Ac) of the account by solving the aforementioned initial value problem. Be sure your solution is correct as this will be used for the remaining parts of this problem. (You can check your answer) (b) Upon graduating from college and starting your career, you have no money. However, during each year after that you deposit d = $1,000 into an account that pays interest at a rate of 8% compounded continuously. Using your solution to part (a), find the equation for the future value A(t) of the account. What is the balance 40 years after you graduate? How much interest is eared on the account after 40 years? Round your answers to the nearest cent. (c) Find the value for an annual deposit d that would produce a balance of one million dollars 40 years later. Round your answer to the nearest cent. (d) If the annual deposit is d = $2,500, what should be the value of the interest r in order for the account balance to be one million dollars after 40 years? You will need a computer to solve this either numerically or graphically. Round your answer to the nearest one-hundredth of a percent.

Answer 1

Differential equation dA = VA+d da = (A+) By Seperation of variables da ardt At of integrating on both sides , dA = (xdt J Ata log (A+ d ) = 8t + log(e) Whele log(e) is an integration Constant We have A (0) = Ao A + d = ceo & c= A + og Therefore [ A(t)= (A0+ flert- &

(b) No money upe o money upon geaduation A=0 d = $ 1000 8=8% = 0.08 A(t) = (0+ 10000.08€ 1000 0.08 0.08 A(+) = 12500 € - 12500) Balance after 40 years: A(40) = 12500 (0.08)(40) J. A(40) = $ 12.500 (23.2.1) A(40) = $ 29 4156.6275 Balance after 40 years is $294157 Deposits made over 40 yeals = 400 = $40,000 Interest earned on the account = A(40) - $40,000 = $ 254157 Interest = $254157

0.08 0 0.08. Act) = de co-ost de A (40) = one million dollar. = $ 1000000 1000000 = de (C0.08) (40),) 08 ( 1 + d = / 80000 1 23.2 1 - $3399.5494 ~ $3400 d- $3400 Alt) = d (ext-1) 9 at += 40 years A (+) = $ 1000000 1000000 = 2500 (408) (6408 ,) = 4008 (8+o) 408-4008-1=0 Let f= 408_4008-1 observe from the plot that too when x=0.09 Therefore interest late - 8=19% of 0.09 |

10 4 0 -154 0 0.01 0.02 0.03 0.07 0.08 0.09 0.1 0.04 0.05 0.06 interest rater