Problem 73 - Calculus an Applied Approach

Question

Compound Interest You deposit \(\$ 1000\) in an account with an annual interest rate of \(r\) (in decimal form) compounded monthly. At the end of 5 years, the balance is $$A=1000\left(1+\frac{r}{12}\right)^{60}.$$ Find the rates of change of \(A\) with respect to \(r\) when (a) \(r=0.08,\) (b) \(r=0.10,\) and (c) \(r=0.12\).

Step-by-Step Solution

Answer

The rates of change are obtained by substituting the given rates into the derivative equation. Hence, these are: (a) \(\frac{dA}{dr} = 5000\left(1 + \frac{0.08}{12}\right)^{59}\) (b) \(\frac{dA}{dr} = 5000\left(1 + \frac{0.10}{12}\right)^{59}\) (c) \(\frac{dA}{dr} = 5000\left(1 + \frac{0.12}{12}\right)^{59}\)

Step 1: Understand the problem

We are given a function for the account balance, \(A\), in terms of the interest rate, \(r\), after a certain period (5 years with monthly compounding). Our task is to find how much \(A\) changes as \(r\) changes, i.e., we need to find \(\frac{dA}{dr}\) for specific values of \(r\).

Step 2: Differentiate the function with respect to \(r\)

Given the relationship \(A = 1000\left(1 + \frac{r}{12}\right)^{60}\), we recognize this as an exponential function. Let \(y = u^{60}\), hence, \(u = 1 + \frac{r}{12}\). The chain rule for differentiation will be used here, with a derivative \(\frac{dy}{dr} = \frac{dy}{du} \times \frac{du}{dr}\). The differentiation gives us: \(\frac{dy}{dr} = 60u^{59} \times \frac{1}{12}\). Subsituting \(u\), the derivative, \(\frac{dA}{dr}\), of \(A\) with respect to \(r\) is: \(\frac{dA}{dr} = 5000\left(1 + \frac{r}{12}\right)^{59}\)

Step 3: Calculate the rate of change at specific values of \(r\)

Now, we substitute the given values of \(r\) from each part of the exercise (0.08, 0.10, 0.12) into the derivative equation to find the rate of change, \(\frac{dA}{dr}\), at these points. Let's do the calculation: (a) \(r=0.08, \frac{dA}{dr} = 5000\left(1 + \frac{0.08}{12}\right)^{59}\) (b) \(r=0.10, \frac{dA}{dr} = 5000\left(1 + \frac{0.10}{12}\right)^{59}\) (c) \(r=0.12, \frac{dA}{dr} = 5000\left(1 + \frac{0.12}{12}\right)^{59}\) These answers will provide the rates of change of the account balance with respect to the interest rate at these specific values.

Key Concepts

Exponential Function

In the context of compound interest, exponential functions play a pivotal role. When you hear the term exponential function, it refers to equations of the form \(a^x\), where \(a\) is a constant base and \(x\) is the exponent.
This type of function is significant in financial calculations because it models the growth of an investment over time.
For compound interest, the equation involves multiple layers of growth because interest is earned on both the initial principal and the accumulated interest from previous periods.
The base of the exponential function for compound interest is \(1 + \frac{r}{n}\), where \(r\) is the interest rate and \(n\) is the number of times interest is compounded per year.

• In our exercise, the compounded base is \(1 + \frac{r}{12}\).
• The exponent, \(60\), comes from 5 years of monthly compounding (12 times per year).

Remember, exponential functions exponentially increase, meaning the function’s rate of increase becomes steeper over time. This reflects how compound interest grows balancing initial investment and reinvested earnings.

Chain Rule

The chain rule is a fundamental concept in calculus, used when differentiating composite functions. When tackling our compound interest function, the expression \(A = 1000\left(1 + \frac{r}{12}\right)^{60}\) is complex, because it's not just a simple polynomial.
We have two nested functions: the base \(1 + \frac{r}{12}\) raised to the exponent, \(60\).
To find the derivative of \(A\) with respect to \(r\), we unravel this using the chain rule. Here's a simplified way to break it down:

• Define \(u = 1 + \frac{r}{12}\).
• The function thus becomes \(A = 1000u^{60}\).

The chain rule tells us to first differentiate the outer function regarding \(u\), which gives \(60u^{59}\), and then multiply by the derivative of the inner function, \(u\), with respect to \(r\) which results in \(\frac{1}{12}\).
Hence, \[\frac{dA}{dr} = 60 \times 1000 \cdot u^{59} \times \frac{1}{12}\] This step helps in identifying the changing nature of \(A\) as \(r\) changes, effectively giving us the sensitivity of the compound interest to variations in the interest rate.

Rate of Change

When we talk about the rate of change in the context of a function, we're discussing how a particular output changes as the input changes.
For our compound interest problem, this means finding out how the account balance \(A\) changes as the interest rate \(r\) changes.
In mathematical terms, this is represented as \(\frac{dA}{dr}\). It tells us 'how fast' or 'slow' \(A\) grows as \(r\) increases.
Since our function is exponential, even small changes in \(r\) can result in significant changes in \(A\).
Given the formula for \(\frac{dA}{dr}\): \[ \frac{dA}{dr} = 5000 \left(1 + \frac{r}{12}\right)^{59} \], We see that:

• Higher values of \(r\) lead to a higher rate of change.
• The exponent \(59\) illustrates the persistence of exponential growth.

This highlights the importance of understanding changes in \(r\) when making financial decisions.
The greater the rate of change, the more sensitive the balance is to fluctuations in the interest rate. Calculating this rate helps in understanding and forecasting the potential growth of investments over time.