Problem 8 - Thomas Calculus in SI Units

Question

Rewrite the expressions in terms of exponentials and simplify the results as much as you can. $$\cosh 3 x-\sinh 3 x$$

Step-by-Step Solution

Answer

The simplified expression is \( e^{-3x} \).

Step 1: Understanding Hyperbolic Functions

The hyperbolic cosine function is given by \( \cosh(x) = \frac{e^x + e^{-x}}{2} \) and the hyperbolic sine function is given by \( \sinh(x) = \frac{e^x - e^{-x}}{2} \). We need to rewrite these in terms of exponentials.

Step 2: Express in Exponentials

We begin by applying the definitions to the given expression \( \cosh(3x) - \sinh(3x) \). This gives us: \[ \cosh(3x) = \frac{e^{3x} + e^{-3x}}{2}, \sinh(3x) = \frac{e^{3x} - e^{-3x}}{2}. \] So the expression becomes: \[ \frac{e^{3x} + e^{-3x}}{2} - \frac{e^{3x} - e^{-3x}}{2}. \]

Step 3: Combine and Simplify

Combine the fractions: \[ \frac{e^{3x} + e^{-3x} - (e^{3x} - e^{-3x})}{2}. \]Simplifying this we have:\[ \frac{e^{3x} + e^{-3x} - e^{3x} + e^{-3x}}{2} = \frac{2e^{-3x}}{2}. \]This further simplifies to: \[ e^{-3x}. \]

Key Concepts

Hyperbolic Identities

Hyperbolic identities are a fundamental aspect of hyperbolic functions, which are analogs to the trigonometric functions but for hyperbolic geometry. The most basic hyperbolic identities involve the hyperbolic sine and cosine functions. These functions are denoted as \( \sinh(x) \) and \( \cosh(x) \) respectively.

The identity for hyperbolic cosine is given by:

• \( \cosh(x) = \frac{e^x + e^{-x}}{2} \)

This formula shows that \( \cosh(x) \) is constructed using exponential functions, making it closely related to exponential growth and decay.

Similarly, the hyperbolic sine function is defined as:

• \( \sinh(x) = \frac{e^x - e^{-x}}{2} \)

These identities help us transform hyperbolic expressions into exponential terms, which makes them easier to simplify and solve in algebraic calculations.

Exponential Functions

Exponential functions are central to many areas of mathematics and science. They describe processes that contract or expand by constant multipliers over equal intervals. This concept is crucial for understanding hyperbolic functions since \( \sinh(x) \) and \( \cosh(x) \) are derived from exponential expressions.

Through this connection, the hyperbolic cosine and sine can be rewritten using the properties of exponentials:

• \( \cosh(x) = \frac{e^x + e^{-x}}{2} \)
• \( \sinh(x) = \frac{e^x - e^{-x}}{2} \)

Consider the expression \( \cosh(3x) - \sinh(3x) \). By substituting the definitions of \( \cosh \) and \( \sinh \), we convert the hyperbolic expression into its exponential form:
\[ \cosh(3x) - \sinh(3x) = \frac{e^{3x} + e^{-3x}}{2} - \frac{e^{3x} - e^{-3x}}{2} \]

Exponential functions are straightforward when dealing with algebraic operations like addition, subtraction, multiplication, and division, and understanding their properties is essential for working with hyperbolic functions.

Simplifying Expressions

Simplifying expressions involves reducing them to their simplest form where they are easiest to work with and understand. In our current context, this means transforming a complex hyperbolic expression into a more manageable exponential function.

Let's break down the simplification process of \( \cosh(3x) - \sinh(3x) \):

• First, write each hyperbolic function in its exponential form.
• Subtract the two expressions by combining like terms.
• Notice that some terms will cancel each other out, specifically the \( e^{3x} \) terms in this case.
• We are left with a simpler expression: \( \frac{2e^{-3x}}{2} \).
• Simplifying gives: \( e^{-3x} \).

By following these steps, you efficiently reduce complex hyperbolic expressions into concise exponential forms, simplifying both the algebra and the interpretation of the solution.