A bank account that accrues continuously compounded interest grows in value at a rate pro- portional to its current value.

Question

The current value is modeled by the exponential equation y = y0e^rt, where r is the annual interest rate, t is time in years, and y0 is the initial value of the account. Suppose after 2 years, the account held $10,000 and after 5 years, the account

held $12,000.

(b) Use this to determine when the account will reach a value of $15,000.

(c) Suppose $10,000 was the “initial” value and $12,000 was the value after 3 years. Find r and the value of the account 2 years BEFORE the “initial” time. In a print statement, explain what you notice when comparing these answers to part (a).

Answer 1

A = yoe^rt

10,000 = yoe^2r

12,000= yoe^5r where yo= initial amount, r=annual interest rate, e=an irrational number about 2.71828

left side = Amount after t years

for continuously compounded interest

divide the last 2 equations above

12/10= 6/5=1.2 = e^(5r-2r) = e^3r

1.2=e^3r

take natural logs of both sides

ln1.2 = 3r

r = (ln1.2)/3 = about .061 = 6.1% rate increase per year

10,000 in 2 years

12,000 in 5 years

increases about 6.1% per year

15000=12000(1.061)^t if 12000 were the initial amount

15/12=5/4=1.25 = 1.061^t

t = log1.25_1.061 = ln1.25/ln1.061 = about 3.769 years from time t=2

add 2

5.769 years to reach $15,000

go backwards 2 years to find yo

yo(1.061^2)=10,000

yo = 10000/(1.061^2)

= about $8,883.20 = initial amount

Answer 2

10,000 = y₀e^2r

12,000 = y₀e^5r

1.2 = e^3r

Use the value of r to answer the following questions.