Problem 18 - Applied Calculus

Question

Simplify the quantities using \(m(z)=z^{2}\). $$m(z+h)-m(z-h)$$

Step-by-Step Solution

Answer

The simplified expression is \(4zh\).

Step 1: Substitute into Function

We need to find the expressions for \(m(z+h)\) and \(m(z-h)\). Since \(m(z) = z^2\), we substitute: - \(m(z+h) = (z+h)^2\)- \(m(z-h) = (z-h)^2\)

Step 2: Expand the Binomials

Next, we expand the binomials:- \((z+h)^2 = z^2 + 2zh + h^2\)- \((z-h)^2 = z^2 - 2zh + h^2\)

Step 3: Substitute Back into the Difference

We substitute the expanded forms back into the expression: \\(m(z+h) - m(z-h) = (z^2 + 2zh + h^2) - (z^2 - 2zh + h^2)\)

Step 4: Simplify the Expression

Combine like terms to simplify:- Start by cancelling out \(z^2\) and \(h^2\), which appear in both terms: \((z^2 + 2zh + h^2) - (z^2 - 2zh + h^2) = 2zh + 2zh\)- Add the remaining terms: \(2zh + 2zh = 4zh\)

Key Concepts

Binomial Expansion

Binomial expansion is a crucial concept in calculus that helps in expanding expressions raised to a power. Understanding it can simplify complex equations. This is particularly helpful in problems like this where expressions like \((z+h)^2\) and \((z-h)^2\) need to be expanded. To start, recognize that a binomial is simply a two-term expression, like \((z+h)\) or \((z-h)\). When such binomials are squared, they expand using the formula:

• \((a+b)^2 = a^2 + 2ab + b^2\)

In our exercise, substituting \(a\) and \(b\) with \(z\) and \(h\), respectively, provides us with:

• \((z+h)^2 = z^2 + 2zh + h^2\)
• \((z-h)^2 = z^2 - 2zh + h^2\)

Exploit this method in all problems where powers and terms are involved. Mastery of binomial expansion streamlines solving and simplifying expressions, providing a firm foundation for more advanced calculus topics.

Simplifying Expressions

Simplifying expressions forms the backbone of solving mathematical problems efficiently. This concept involves collecting like terms to reduce an expression to its simplest form. In the given exercise, to simplify the expression \(m(z+h) - m(z-h)\), we begin with the expanded forms of each function output, \((z^2 + 2zh + h^2)\) and \((z^2 - 2zh + h^2)\). The key is to

• Identify like terms: both expressions have \(z^2\) and \(h^2\), which directly cancel each other out.
• Focus on terms that undergo changes, such as \(2zh\), by evaluating them separately.

Thus, subtracting these forms results in:
\((z^2 + 2zh + h^2) - (z^2 - 2zh + h^2) = 4zh\)
What remains is the simplified product \(4zh\). Learning to simplify deftly is a skill that will aid in calculus and beyond, allowing for verification of problems more intuitively.

Difference Quotient

The concept of the difference quotient is central to understanding derivatives in calculus. It provides a way to represent average rates of change over an interval, and ultimately, the slope of the tangent line at a point. Derived from a slightly different context, the exercise shows how expressions can be altered before finding such quotients.
Typically, the difference quotient takes the form:

• \[\frac{f(x+h) - f(x)}{h}\]

However, problems may involve modified scenarios, akin to finding equivalents like \(m(z+h) - m(z-h)\). While the current exercise does not explicitly require division by \(h\), understanding its reduction still embodies the spirit of the difference quotient.
Realizing how these expressions simplify before reaching the quotient stage prepares students for handling limits,which is the gateway to finding derivatives.