If an investment of d dollars is compounded continuously at an annual interest r
ID: 3345094 • Letter: I
Question
If an investment of d dollars is compounded continuously at an annual interest rate r, then the amount of money A(t) earned at time t>0(in years) is the solution of the initial value problem A'(t)=rA(t) A(0)=d where r should be expressed as a fraction rather than as a percentage rate, e.g., an annual interest rate of r=12% would be expressed as r=12/100. Suppose you have just placed d dollars in a bank account that pays 1% annual interest, compound continuously. How much do you have in the account after 5 years? How long(in years) will it take your money to double? How long(in years) will it take your money to triple?Explanation / Answer
We are given: A'(t) = rA(t) ----- Rewrite this as:
dA/dt = rA
1/A dA = r dt
Integrating boths sides gives us:
ln |A| = rt + c
A = e^(rt+c) ------ let p = e^c
A = pe^(rt)
Our initial value is A(0) = d, and r=.01 so:
d=pe^(.01 * 0)
d=pe^0
d=p
This gives us the formula:
A=de^(rt) where A is the end amount, d is the dollars placed in the bank, t is the length of time
How much money after 5 years?
A = d*e^(.01*5)
A.= d*e^(.05)
A = d * 1.05
You will have 1.05 times as much money as you put in
How long does it take to double your money?
2d=d*e^(.01*t)
2=e^(.01*t)
ln(2) = .01*t
t = ln(2)/.01
t = 69.31
How long does it take to triple your money?
3d=de^(.01*t)
3=e^(.01*t)
ln(3) =.01*t
t = ln(3)/.01
t = 109.86