Problem 9 - Calculus an Applied Approach

Question

Find the function values. $$ A(P, r, t)=P\left[\left(1+\frac{r}{12}\right)^{12 t}-1\right]\left(1+\frac{12}{r}\right) $$ $$ \begin{array}{llll}{\text { (a) } A(100,0.10,10)} & {\text { (b) } A(275,0.0925,40)} & {} & {}\end{array} $$

Step-by-Step Solution

Answer

The function values are approximately 13150 for \( A(100, 0.10, 10) \) and 1200372.75 for \( A(275, 0.0925, 40) \).

Step 1: Substituting the first set of values into the equation

Substitute \( P = 100 \), \( r = 0.10 \), and \( t = 10 \) into the equation. The function becomes: \[ A(100, 0.10, 10) = 100 \left[ \left(1+ \frac{0.10}{12}\right)^{12*10} - 1 \right] \left(1+ \frac{12}{0.10}\right)\]

Step 2: Calculate the first function value

Calculate the value of the function, which is an expression involving mathematical operations of addition, multiplication, division, and exponentiation. The result is approximately 13150.

Step 3: Substituting the second set of values into the equation

Next, substitute \( P = 275 \), \( r = 0.0925 \), and \( t = 40 \) into the equation. The function becomes: \[ A(275, 0.0925, 40) = 275 \left[ \left(1+ \frac{0.0925}{12}\right)^{12*40} - 1 \right] \left(1+ \frac{12}{0.0925}\right) \]

Step 4: Calculate the second function value

Again perform the required calculations. The result is approximately 1200372.75.

Key Concepts

Financial Functions

When we talk about financial functions, we are referring to mathematical equations that help determine financial outcomes. In this exercise, we use a financial function to calculate compound interest. Understanding this function is important because it shows how money can grow over time when it is invested or borrowed.

To find out how much money you would have after a certain period, you plug in three main components:

• Principal amount \( P \) - the starting amount of money.
• Interest rate \( r \) - the rate at which the money will grow, usually expressed as a decimal.
• Time \( t \) - the time period in years for which the money will be invested or borrowed.
• Compounding frequency - here it is monthly, which is a crucial part for calculating compound interest accurately.

These components work together in an equation to reflect the growth of your investment over the years. This makes it a powerful tool in financial planning and understanding future values of investments.

Mathematical Operations

Mathematical operations are the building blocks of solving financial functions. They include addition, subtraction, multiplication, division, and exponentiation. Each of these operations plays a crucial role in calculating the values of financial functions.

In our compound interest formula, we need to use:

• **Addition:** Where we add \( rac{r}{12} \) to 1 to get the multiplier in the bracket.
• **Division:** Used to break down annual interest into monthly terms, utilizing \( rac{r}{12} \).
• **Multiplication:** Several parts of this equation involve multiplication, like multiplying the base interest calculation by 12 for compounding monthly and then multiplying with principal \( P \).
• **Exponentiation:** Needed to compound the interest over time \( t \), this operation explains the effect of compounded interest.

Taking the time to understand each operation ensures accurate calculations and helps build confidence in dealing with financial equations.

Exponentiation in Algebra

Exponentiation is a mathematical operation involving numbers raised to a power. It is essential in calculating compound interest because it reflects how interest is applied multiple times over a period.

In our formula, the operation \( \left(1 + \frac{r}{12}\right)^{12t} \) is an exponentiation expression. Here you:

• **Raise the monthly compound interest factor to the power of the total number of compounding periods.** Since interest compounds monthly, the expression \( 12t \) calculates total compounding instances.

This repeated multiplication of the base \( \left(1 + \frac{r}{12}\right) \) each period expresses how investment grows exponentially over time. Instead of a simple additions of interest, exponentiation dominates the growth pattern, leading to substantial increases which are characteristic of compound growth. When calculating it manually, ensure meticulous care especially in multi-step algebra problems to derive correct solutions.