Problem 19 - Applied Calculus

Question

Simplify the quantities in Problems \(16-19\) using \(m(z)=z^{2}\) $$m(z+h)-m(z-h)$$

Step-by-Step Solution

Answer

The simplified expression is \(4zh\).

Step 1: Understanding the Function

The given function is \( m(z) = z^2 \). We need to find the difference \( m(z+h) - m(z-h) \). This means we need to substitute \( z+h \) and \( z-h \) into the function \( m(z) \), which is \( z^2 \).

Step 2: Substituting and Expanding

Substitute \( z+h \) into the function: \( m(z+h) = (z+h)^2 = z^2 + 2zh + h^2 \). Now, substitute \( z-h \) into the function: \( m(z-h) = (z-h)^2 = z^2 - 2zh + h^2 \).

Step 3: Finding the Difference

Now compute the difference: \( m(z+h) - m(z-h) = (z^2 + 2zh + h^2) - (z^2 - 2zh + h^2) \). Simplify this expression by combining like terms.

Step 4: Simplifying the Expression

In the difference \((z^2 + 2zh + h^2) - (z^2 - 2zh + h^2)\), the \(z^2\) terms and \(h^2\) terms cancel each other out. This leaves: \( 2zh + 2zh = 4zh \).

Key Concepts

Function Expansion

When dealing with function expansion, we explore how functions change when their inputs are varied. This process involves substituting the input variables with expressions that include additional terms. For the given function, \( m(z) = z^2 \), we expand by substituting inputs. For instance:

• Substitute \( z+h \): \( m(z+h) = (z+h)^2 \)
• Substitute \( z-h \): \( m(z-h) = (z-h)^2 \)

This substitution leads to the expansion of the function into their algebraic expressions. The aim is to find out how changes in input, like differences \( h \), influence the output. Understanding this concept helps in exploring function behaviors, crucial in fields like calculus where derivatives, which measure such changes, play a vital role.
Function expansion allows us to analyze the underlying algebraic properties of a function when subjected to different input alterations.

Algebraic Expression Simplification

Simplifying algebraic expressions involves condensing complex expressions into simpler forms without changing their values. This process often includes combining like terms or reducing fractions. To understand the simplification process:
Consider the expression \((z^2 + 2zh + h^2) - (z^2 - 2zh + h^2)\) from the solution. By subtracting the second expression from the first, we encounter terms that naturally cancel out:

• \(z^2\) terms cancel with \(z^2\)
• Positive \(h^2\) cancels with negative \(h^2\)

This results in the simplification to \(4zh\).
The art of simplifying revolves around recognizing and eliminating terms that offset each other. Such simplifications help make calculations more manageable and reveal the essential parts of an expression.

Quadratic Functions

Quadratic functions are polynomial functions of degree two, which take the general form \(m(z) = az^2 + bz + c\). In our exercise, we specifically deal with the simplest form \(m(z) = z^2\), where \(a=1\), and both \(b\) and \(c\) are zero.These functions graph as parabolas, symmetrical around their vertex, and are fundamental in algebra. Quadratic functions are crucial for many applications:

• In physics, they describe trajectories
• In finance, they model certain economic behaviors

When we examine function expansions as seen by inputting \(m(z+h)\) and \(m(z-h)\), these inputs allow us to determine the function's behavior around a specific point. Understanding quadratics and their properties lays a foundation for exploring more advanced mathematics, including calculus and real-world problem solving.