In compound interest, we would have heard the terms like 'compounded annually', 'compounded semi annually', 'compounded quarterly' and 'compounded monthly'.
But, always we have question about compounded continuously. To understand 'compounded continuously', let us consider the example given below.
When we invest some money in a bank, it will grow continuously. That is, at any instant the balance is changing at a rate that equals 'r' (rate of interest per year) times the current balance.
Formula for compound interest :
A(t) = P(1 + r/n)^{t}
A ---> Final value
P ---> Initial value
r ---> Growth rate (in percent)
t ---> Time (in years)
n ---> number of compounding periods
In the compound interest formula A(t) = P(1 + r/n)^{t},
Annual compounding
1 time in a year
n = 1
Half yearly compounding
2 times in a year
n = 2
Quarterly compounding
4 times in a year
n = 4
Monthly compounding
12 times in a year
n = 12
Daily compounding
365 times in a year
n = 365
Continuous compounding
Infinite number of times in a year
n ---> ∞
Many real world phenomena are being modeled by functions which describe how things grow continuously at any instance.
Suppose we want to compound continuously. This means that the number of compounding periods n grows without bound, i.e., n --->∞.
So, we have
The formula given below is related to compound interest formula and represents the case where interest is being compounded continuously.
A = Pe^{rt}
A ---> Final value
P ---> Initial value
r ---> Growth rate (in percent)
t ---> Time (in years)
Example 1 :
You invest $2500 in bank which pays 10% interest per year compounded continuously. What will be the value of the investment after 10 years ?
Solution :
Formula for continuous compounding :
A = Pe^{rt}
Substitute P = 2500, r = 10% or 0.1, t = 10, e ≈ 2.71828.
A = 2500(2.71828)^{0.1 x 10}
= 2500(2.71828)^{1}
= $6795.70
Example 2 :
If David invests $500 at annual rate of 20% compounded continuously, calculate the final amount that David will have after 5 years.
Solution :
Formula for continuous compounding :
A = Pe^{rt}
Substitute P = 500, r = 20% or 0.2, t = 5, e ≈ 2.71828.
A = 500(2.71828)^{0.2 x 5}
= 500(2.71828)^{1}
= $1359.14
Kindly mail your feedback to v4formath@gmail.com
We always appreciate your feedback.
©All rights reserved. onlinemath4all.com
Jan 19, 22 09:34 AM
Multiplicative Inverse Worksheet
Jan 19, 22 09:25 AM
Multiplicative Inverse - Concept - Examples
Jan 19, 22 08:24 AM
Graphing Linear Functions Worksheet