Question

Level curves of a savings account Suppose you make a one-time deposit of \(P\) dollars into a savings account that earns interest at an annual rate of \(p \%\) compounded continuously. The balance in the account after \(t\) years is \(B(P, r, t)=P e^{r t},\) where \(r=p / 100\) (for example, if the annual interest rate is \(4 \%,\) then \(r=0.04\) ). Let the interest rate be fixed at \(r=0.04.\) a. With a target balance of \(\$ 2000,\) find the set of all points \((P, t)\) that satisfy \(B=2000 .\) This curve gives all deposits \(P\) and times \(t\) that result in a balance of \(\$ 2000\). b. Repeat part (a) with \(B=\$ 500, \$ 1000, \$ 1500,\) and \(\$ 2500\) and draw the resulting level curves of the balance function. c. In general, on one level curve, if \(t\) increases, does \(P\) increase or decrease?

Short answer

Answer: As the time (t) increases, the initial deposit (P) needed to reach a particular target balance decreases.

Step-by-step solution

(a) Set of all points for a balance of \(\$ 2000\)#

Let's substitute the target balance \(B = 2000\) and interest rate \(r = 0.04\) into the function \(B(P, r, t) = P e^{r t}\) and solve for \(P\) as a function of \(t\), as follows \(2000 = P e^{0.04 t}\) Now let's solve it for \(P\): \(P = \frac{2000}{e^{0.04 t}}\) Thus, any \((P, t)\) that satisfies this equation will result in a balance of \(\$ 2000\).

(b) Level curves for other balances#

Following the same procedure, we can find the equations for the level curves of balances \(\$ 500\), \(\$ 1000\), \(\$ 1500\), and \(\$ 2500\): (i) For \(B = 500\): \(P = \frac{500}{e^{0.04 t}}\) (ii) For \(B = 1000\): \(P = \frac{1000}{e^{0.04 t}}\) (iii) For \(B = 1500\): \(P = \frac{1500}{e^{0.04 t}}\) (iv) For \(B = 2500\): \(P = \frac{2500}{e^{0.04 t}}\) Now let's draw the level curves for each balance: 1. On the horizontal axis, label time (\(t\)) in years. 2. On the vertical axis, label the initial deposit (\(P\)) in dollars. 3. Draw the curve for each balance using the equations (i) to (iv) above. 4. Label each curve with its corresponding balance.

(c) Increase or decrease of \(P\) as \(t\) changes#

In general, as we move along one level curve, we see that as \(t\) increases, \(P\) decreases. This trend can be explained by the compound interest formula. As time goes by, a smaller initial deposit is needed to reach a particular target balance due to the effect of compound interest.

Key concepts

Level Curves

Level curves are an excellent way to visualize different possibilities for reaching the same target balance in a savings account over time. They illustrate the relationship between two variables, in this case, the initial deposit amount \(P\) and the time \(t\). Each curve represents combinations of \(P\) and \(t\) that result in the same final balance \(B\). For instance, with a fixed interest rate, you may achieve a balance of \(2000 by either depositing more money initially or waiting longer if your initial deposit is smaller.

• Each level curve indicates a different target balance, like \)500, \(1000, \)1500, etc.
• On a graph, \(P\) is plotted on the vertical axis (initial deposit), and \(t\) is on the horizontal axis (time in years).
• The curves tend to slope downward, indicating that a smaller initial deposit is needed as the time increases, due to the effects of compound interest.

Level curves are useful for financial planning as they allow you to assess how much you need to deposit today to reach a future financial goal, taking into account the time you have and the interest rate available.

Exponential Growth

Exponential growth is a fundamental concept in understanding how balances in a savings account grow over time. In a bank account with continuously compounded interest, the balance grows faster as time progresses.

• This is expressed using the formula \(B = P e^{rt}\), where \(B\) is the final balance, \(P\) is the initial deposit, \(r\) is the annual interest rate (in decimal form), and \(t\) is time in years.
• The exponential function \(e^{rt}\) captures how the rate of increase is proportional to the current amount.
• This means that not only the initial deposit grows, but the interest earned grows at a rate that compounds over time, leading to a rapid increase.

Exponential growth is critical in financial contexts because it illustrates how investments and savings can grow dramatically with time, especially when left untouched for longer durations. It emphasizes the power of letting interest do its work for you.

Compound Interest

Compound interest is the process by which interest is added to the principal sum of a deposit, and then in subsequent periods, interest is earned on the new total. This effect of earning "interest on interest" is what differentiates compound interest from simple interest.

• It's calculated using the formula \(A = P\left(1 + \frac{r}{n}\right)^{nt}\), where \(A\) is the amount of money accumulated after \(n\) years, including interest.
• \(P\) is the principal amount, \(r\) is the annual interest rate (decimal), \(n\) is the number of times interest is compounded per year, and \(t\) is the time the money is invested for, in years.
• Unlike simple interest, which only earns interest on the initial amount, compound interest continually earns on itself, thus accelerating growth.

Understanding compound interest is crucial to make informed financial decisions, as it can significantly impact the growth of both savings and the cost of loans.

Continuously Compounded Interest

Continuously compounded interest is a limiting case of compound interest, where interest is compounded an infinite number of times per year.

• The formula to calculate continuously compounded interest is \(A = P e^{rt}\), where \(A\) is the amount of money accumulated after \(t\) years, \(P\) is the principal amount, \(r\) is the annual interest rate (as a decimal), and \(e\) is Euler's number (approximately 2.71828).
• This method of compounding assumes that the interest is being theoretically added continuously, resulting in the maximum possible amount of growth for a given rate and period.
• It is commonly used in theoretical calculations because of its simplicity and elegance, despite not always being implemented in real-life financial products.

Continuously compounded interest provides the upper bound for growth through interest and is particularly significant for those studying or working in fields needing precise and optimized growth models, such as finance or physics.