Problem 23 - Calculus: An Integrated Approach to Functions and Their Rates of Change

Question

(a) You plan to save money starting today at a rate of \(\$ 4000\) per year over the next 30 years. You will deposit this money at a nearly continuous rate (a constant amount each day) into a bank account that earns \(5 \%\) interest compounded continuously. Let \(B(t)\) be the balance of money in the account \(t\) years from now, where \(0 \leq t \leq 30\) i. Write a differential equation whose solution is \(B(t)\). ii. Write an integral that is equal to \(B(30)\), the amount in the account at the end of 30 years. (b) Now assume that instead of making deposits continuously, you decide to make a deposit of \(\$ 4000\) once a year, starting today and continuing until you have made a total of 30 deposits. Suppose the bank account pays \(5 \%\) interest compounded annually. i. Write a geometric sum equal to the balance immediately after the final deposit. ii. Find a closed form expression (no \(+\cdots+\), no summation notation) for this sum.

Step-by-Step Solution

Answer

The differential equation that represents continuous deposits is \(B'(t) = 0.05B(t) + 4000\), and the integral expression that represents the amount in the account after 30 years is \(B(30) = \int_{0}^{30} (0.05B(t) + 4000) dt \). The geometric sum that represents annual deposits is [\(4000(1+0.05)^{30} + 4000(1+0.05)^{29} +...+ 4000(1+0.05)^{1} + 4000\)] and the closed form of this sum is \(4000[(1.05^{30} -1)/0.05]\).

Step 1: Continuous deposit- Differential Equation

The given scenario described continuous deposit of money and can be put into a differential equation. It would be as follows: \(B'(t) = 0.05B(t) + 4000\). This is because the rate of increase in the balance is not only dependent on the current balance (first term on the right), but also on the continuous deposits (second term on the right).

Step 2: Amount at the end of 30 years - Integral Expression

The balance after 30 years, \(B(30)\), is the integral of the rate of increase of the balance over the interval \([0, 30]\). So, we can get the integral expression as : \(B(30) = \int_{0}^{30} (0.05B(t) + 4000) dt \). Remember, since \(B'(t)\) is not just a function of \(t\), but also of \(B(t)\), this is not a straightforward integral and usually solving such equations requires methods in differential equations.

Step 3: Annual deposit- Geometric Series

In case of considering annual depositing of money, the balance immediately after the final deposit is a geometric series where each term is the deposit made that year grown with interest over the remaining years. We can write the balance after 30 years as \([4000(1+0.05)^{30} + 4000(1+0.05)^{29} +...+ 4000(1+0.05)^{1} + 4000]\). Each term is the deposit (\$4000) multiplied with the interest grown over the remaining years.

Step 4: Closed Form of the Geometric Series

The geometric series can be expressed in a closed form using the formula for the sum of a geometric series. The sum of a geometric series where \(a\) is the first term, \(r\) is the common ratio and \(n\) is the number of terms, is given by \([a(r^n -1)] / (r-1)\). For the given series, \(a=4000\), \(r=(1+0.05)=1.05\) and \(n=30\). So, the sum \(= 4000[(1.05^{30} -1)/0.05]\).

Key Concepts

Differential Equations

Differential equations are a powerful tool in mathematics and are used to describe how quantities change over time. In the exercise above, we are asked to create a differential equation to model the balance of an account with continuous deposits. A differential equation generally involves an unknown function and its derivatives.

In our scenario, the balance in the bank account, represented by \(B(t)\), changes because of two factors:

• Growth due to interest: As the balance \(B(t)\) continuously accrues interest at a rate of 5%, its contribution to the balance's rate of change is \(0.05B(t)\).
• Continuous deposits: The continuous deposit of \$4000 per year adds \(4000\) to the rate of change of the balance.

Thus, the differential equation modeling the scenario is \(B'(t) = 0.05B(t) + 4000\). Solving this type of differential equation, known as a first-order linear differential equation, typically involves techniques like integrating factors or separation of variables. However, the integration step may not be straightforward because the rate of change is dependent on the current balance, making the math more involved.

Ultimately, this differential equation captures the dynamic nature of the continuous deposit process combined with compound interest over time.

Geometric Series

A geometric series is a series of numbers where each term after the first is obtained by multiplying the previous term by a fixed, non-zero number called the common ratio. When saving money annually, as described in part (b) of the exercise, the balance relies on a geometric series.

Here, each deposit of \$4000 earns interest over the remaining years until the end of 30 years. The contribution of each deposit compounds annually. For instance, the first deposit grows for 30 years, while the last deposit grows for only 1 year.

• First term \(a = 4000 \times (1.05)^{30}\), assuming a 5% interest rate.
• Each subsequent term is a product of the previous one by \((1.05)\), with the common ratio \(r = 1.05\).
• The number of terms \(n = 30\) represents each year's deposit.

The series becomes \([4000(1.05)^{30} + 4000(1.05)^{29} + \ldots + 4000]\).

To find the total balance after the final deposit using this series, a closed form expression can simplify the process. The formula for the sum of a geometric series is \(S_n = a \frac{(r^n - 1)}{r - 1}\), where the sum \(S_n\) gives us the total value of all deposits compounded over time.

Compound Interest

Compound interest is the process where the interest earned on an account is added back to the principal, so from that point onwards, the interest is calculated on the new, larger principal. This results in an exponential growth of the balance over time.

In the context of this exercise, we examined two types of compound interest:

• Continuous compounding: With continuous deposits, interest is compounded continuously. The formula used is \(A = Pe^{rt}\), where \(P\) is the initial principal, \(r\) is the annual interest rate, \(t\) is the time in years, and \(e\) is the base of the natural logarithms. This gives a smooth, uninterrupted growth curve for the balance.
• Annual compounding: When deposits are made annually, the interest is compounded once per year. The balance grows according to each deposit growing for the number of years left until 30 years due to the compounding effect.

Each method of compounding affects the total amount differently due to the timing and frequency of how interest is applied to the principal. Compound interest, both continuous and discrete (annual in this case), exploits the principle of earning 'interest on interest', leading to larger account balances over extended periods.