Problem 363 - Calculus Volume 1

Question

For the following exercises, use \(y=y_{0} e^{k t}\). What continuous interest rate has the same yield as an annual rate of \(9 \%\) ?

Step-by-Step Solution

Answer

The equivalent continuous interest rate is approximately 8.61777%.

Step 1: Understand the Problem Statement

We are given two interest rates: an annual rate of 9% and a continuous interest rate that needs to have the same yield after 1 year. We will use the equation \(y = y_0 e^{kt}\) which applies to continuous growth scenarios to find this equivalent continuous rate \(k\).

Step 2: Express the Annual Rate as an Equation

When compounded annually, an interest rate of 9% grows by a factor of \(1 + 0.09 = 1.09\) in one year. Hence, \(y = y_0 \cdot 1.09\).

Step 3: Express the Continuous Rate as an Equation

Using the formula for continuous compounding, after 1 year the growth equation is \(y = y_0 e^{k \cdot 1} = y_0 e^k\).

Step 4: Equate the Two Equations

We want the yield of the continuous rate to match the annual compounding rate. So, set the two equations equal: \(y_0 e^k = y_0 \cdot 1.09\). Cancel \(y_0\) from both sides, leading to \(e^k = 1.09\).

Step 5: Solve for the Continuous Rate \(k\)

Take the natural logarithm on both sides to solve for \(k\): \[k = \ln(1.09)\].

Step 6: Calculate \(k\)

Using a calculator, find the value of \(k\): \[k \approx \ln(1.09) \approx 0.0861777\].

Step 7: Interpretation of Result

This means that a continuous interest rate of approximately \(8.61777\%\) yields the same as a 9\% annual interest rate.

Key Concepts

Exponential Growth Equation

Exponential growth is a process where the rate of growth is proportional to the current value, leading to growth with an accelerating rate. One commonly used equation to describe exponential growth is \( y = y_0 e^{kt} \), where:

• \( y \) is the final amount after time \( t \)
• \( y_0 \) is the initial amount
• \( k \) is the growth rate
• \( t \) is the time period

The equation is essential in problems involving continuous growth, like population growth or certain financial investments. In our exercise, this equation was used to find an equivalent continuous interest rate. The attractiveness of exponential growth lies in its perpetual compounding effect, often leading to significant increases over time.

Compound Interest

Compound interest is the interest on a loan or deposit, calculated based on both the initial principal and the accumulated interest from previous periods. It contrasts with simple interest, which is calculated only on the principal amount. The formula for compound interest is usually \( A = P(1 + r/n)^{nt} \), where:

• \( A \) is the future value of the investment/loan
• \( P \) is the principal investment amount
• \( r \) is the annual interest rate (decimal)
• \( n \) is the number of times interest is compounded per year
• \( t \) is the number of years the money is invested for

In the problem, we compared a situation with continuous compounding to annual compounding. Annual compounding implies the interest is applied once per year, while continuous compounding continuously adds interest, calculated as \( A = Pe^{rt} \). Continuous compounding is a special case that treats the compounding process as taking place at every possible instant, leveraging Euler's number \( e \).

Natural Logarithm

The natural logarithm, denoted as \( \ln \), is a fundamental concept in mathematics, particularly useful in solving exponential equations. It is the inverse operation of exponentiation when the base is Euler's number \( e \). For instance, if \( e^x = a \), then \( \ln(a) = x \).

In the solution to our exercise, the natural logarithm was used to "undo" the exponential part of the equation. When we needed to solve \( e^k = 1.09 \) for \( k \), we took the natural logarithm of both sides, yielding \( k = \ln(1.09) \). This operation helps in solving for the exponent in equations where it's initially difficult to isolate \( k \).

The natural logarithm is particularly handy in calculus and various growth models, as it simplifies expressions involving exponentials and provides exact values for continuous compounding scenarios.