Problem 6
Question
A person deposits money into an account, which pays \(5 \%\) interest compounded continuously, at a rate of \(\$ 1000\) per year for 15 years. Calculate: (a) The balance in the account at the end of the 15 years. (b) The amount of money actually deposited into the account. (c) The interest earned during the 15 years.
Step-by-Step Solution
Verified Answer
(a) $28,316.20, (b) $15,000, (c) $13,316.20.
1Step 1: Understanding Compound Interest
The problem involves money deposited continuously with interest compounding continuously. The formula for the future value of continuous compounding is given by the integral \(A = \int_{0}^{T} P \, e^{r(T-t)} \, dt\) where \(P\) is the rate of deposit, \(r\) is the interest rate, and \(T\) is the total time.
2Step 2: Applying the Formula
Substitute the given values into the formula: the annual deposit \(P = 1000\), interest rate \(r = 0.05\) (5%), and time \(T = 15\) years. This gives the integral \[A = \int_{0}^{15} 1000 \, e^{0.05(15-t)} \, dt.\]
3Step 3: Evaluating the Integral
To solve the integral, find the antiderivative: \[A = 1000 \left[ -\frac{1}{0.05} e^{0.05(15-t)} \right]_0^{15}.\] This simplifies to \[A = -20000 \left[ e^{0} - e^{0.05 \times 15} \right].\]
4Step 4: Calculating the Future Value
Evaluate the expression: \[A = -20000 (1 - e^{0.75}).\] Calculate e and subtract: \(e^{0.75} \approx 2.117,\) so \[A \approx -20000 (1 - 2.117) \approx 28316.20.\] Thus, the balance at the end of 15 years is approximately \$28,316.20\.
5Step 5: Calculating Money Deposited
To find the total amount deposited, multiply the rate per year by the number of years: \[\text{Total Deposited} = 1000 \times 15 = 15000.\] This means \$15,000\ was actually deposited over the 15 years.
6Step 6: Calculating Interest Earned
Subtract the total money deposited from the future value to find the interest earned: \[\text{Interest} = 28316.20 - 15000 = 13316.20.\] Therefore, the interest earned over 15 years is approximately \$13,316.20\.
Key Concepts
Compound Interest FormulaIntegral CalculationFuture Value
Compound Interest Formula
Continuous compounding is a way to calculate interest that accrues continuously, providing a very precise and realistic method of interest calculation. Unlike simple or discrete compounding, where interest is calculated periodically, continuous compounding assumes that interest is added an infinite number of times. This is particularly useful for modeling scenarios where deposits occur gradually over time, such as in savings or investment accounts.
The formula to calculate the future value of an account with continuous compounding is derived from the integral calculation. It is given by \[A = \int_{0}^{T} P \, e^{r(T-t)} \, dt\]where:
The formula to calculate the future value of an account with continuous compounding is derived from the integral calculation. It is given by \[A = \int_{0}^{T} P \, e^{r(T-t)} \, dt\]where:
- \(A\) is the future balance of the account,
- \(P\) is the rate of money deposited per year,
- \(r\) is the continuous interest rate (expressed as a decimal),
- \(T\) is the total time period in years.
Integral Calculation
To derive the future value of a continuous compounding scenario, we understand that integral calculation is a crucial technique. By setting up the integral expression \[A = \int_{0}^{15} 1000 \, e^{0.05(15-t)} \, dt,\]we aim to sum up the values of continuous deposits compounded at the interest rate \(0.05\) over 15 years.
Evaluating this integral starts with finding an appropriate antiderivative. The specific antiderivative is dependent on the exponential function within the integral, which involves the compound interest rate. The solution, in this case, \[A = 1000 \left[ -\frac{1}{0.05} e^{0.05(15-t)} \right]_0^{15},\]provides a way to capture the entire continuous compounding behavior.
The integral is then solved by substituting the limits, leading to the final expression \[A = -20000 (1 - e^{0.75}).\]This approach allows us to break down continuous growth into calculable segments and sum these effectively.
Evaluating this integral starts with finding an appropriate antiderivative. The specific antiderivative is dependent on the exponential function within the integral, which involves the compound interest rate. The solution, in this case, \[A = 1000 \left[ -\frac{1}{0.05} e^{0.05(15-t)} \right]_0^{15},\]provides a way to capture the entire continuous compounding behavior.
The integral is then solved by substituting the limits, leading to the final expression \[A = -20000 (1 - e^{0.75}).\]This approach allows us to break down continuous growth into calculable segments and sum these effectively.
Future Value
The future value in the context of continuous compounding refers to the total balance expected at the end of the investment period when deposits have continuously grown. Here, it was calculated to be approximately \(28,316.20 at the end of 15 years as derived from our integral calculation. This future value considers the power of compounding, where interest earns interest, leading to exponential growth.
Future value calculations help in identifying both the investment's performance and the efficiency of the compounding process. In our example, the total of \\)15,000 was initially deposited, growing significantly due to the continuous interest at a spot rate of 5% compounded over those years.
Contrast this with the interest earned, which represents the growth beyond simply adding deposits. Interest valued at around \$13,316.20 was obtained purely from the compounding process. This differentiation is crucial for investors understanding their returns and planning their financial strategies effectively.
Future value calculations help in identifying both the investment's performance and the efficiency of the compounding process. In our example, the total of \\)15,000 was initially deposited, growing significantly due to the continuous interest at a spot rate of 5% compounded over those years.
Contrast this with the interest earned, which represents the growth beyond simply adding deposits. Interest valued at around \$13,316.20 was obtained purely from the compounding process. This differentiation is crucial for investors understanding their returns and planning their financial strategies effectively.
Other exercises in this chapter
Problem 6
Using the Fundamental Theorem, evaluate the definite integrals in Problems \(1-20\) exactly. $$\int_{1}^{2} \frac{1}{x} d x$$
View solution Problem 6
Find the integrals. $$\int y \sqrt{y+3} d y$$
View solution Problem 6
Decide if the function is an antiderivative of \(f(x)=2 e^{2 x}\) $$F(x)=e^{2 x}+\int_{0}^{1} e^{2 t} d t$$
View solution Problem 7
Find the integrals .Check your answers by differentiation. $$\int \frac{x}{\sqrt{x^{2}+4}} d x$$
View solution