Problem 434 - Calculus Volume 1

Question

For the following exercises, solve each problem. prove the expression for \(\cosh ^{-1}(x)\) . Multiply \(x=\cosh (y)=(1 / 2)\left(e^{y}-e^{-y}\right)\) by 2\(e^{y}\) and solve for \(y\) Does your expression match the textbook?

Step-by-Step Solution

Answer

The derived expression matches the textbook: \( \cosh ^{-1}(x) = \ln(x + \sqrt{x^2 - 1}) \).

Step 1: Define Hyperbolic Cosine

The hyperbolic cosine function is defined as \( \cosh(y) = \frac{1}{2} \left(e^{y} + e^{-y}\right) \). To find \( y \) in terms of \( x \), we start with this equation and set \( x = \cosh(y) \), so \( x = \frac{1}{2} \left(e^{y} + e^{-y}\right) \).

Step 2: Isolate Exponential Terms

Multiply both sides by 2 to eliminate the fraction: \( 2x = e^{y} + e^{-y} \). Then, multiply the entire equation by \( e^{y} \) to clear the denominator of \( e^{-y} \): \( 2xe^{y} = e^{2y} + 1 \).

Step 3: Rearrange to Solve for Exponential

Rearrange the equation to form a quadratic in terms of \( e^{y} \): \( e^{2y} - 2xe^{y} + 1 = 0 \). This expression is a quadratic equation in \( e^{y} \).

Step 4: Apply Quadratic Formula

Use the quadratic formula \( e^{y} = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 1, b = -2x, c = 1 \). Substituting these into the quadratic formula gives: \( e^{y} = \frac{2x \pm \sqrt{4x^2 - 4}}{2} \).

Step 5: Simplify the Expression

Simplify the solutions: \( e^{y} = x \pm \sqrt{x^2 - 1} \). Choose the positive branch because \( e^{y} \) is always positive, thus \( e^{y} = x + \sqrt{x^2 - 1} \).

Step 6: Solve for y

Take the natural logarithm on both sides to solve for \( y \): \( y = \ln(x + \sqrt{x^2 - 1}) \). This is the standard expression for the inverse hyperbolic cosine: \( \cosh^{-1}(x) = \ln(x + \sqrt{x^2 - 1}) \).

Key Concepts

Hyperbolic Cosine

The hyperbolic cosine function, often denoted as \( \cosh(x) \), is a close cousin to the regular cosine function, but it involves exponential growth. It is defined by the formula: \[\cosh(y) = \frac{1}{2} (e^{y} + e^{-y})\]Unlike its trigonometric counterpart, hyperbolic cosine is crucial in areas involving hyperbolic geometry and has properties that make it applicable in hyperbolic equations.

• Hyperbolic functions relate to the shape and properties of a hyperbola, similar to how trigonometric functions relate to a circle.
• In the definition, \( e^{y} \) represents the exponential function, a key player in various fields of mathematical study.

Although \(\cosh(y)\) might appear complex, it exhibits symmetry and unique growth patterns, becoming quite intuitive once you see it in action. Specifically, the function opens up the solution space for problems involving rates of hyperbolic change, like some types of population growth or architectural designs.The expression for the inverse hyperbolic cosine, \( \cosh^{-1}(x) \), aims to find a value \( y \) such that \( \cosh(y) = x \). This gives a real-world application of the formula, particularly in understanding hyperbolic arcs and angles.

Quadratic Formula

The quadratic formula is a powerful tool for solving quadratic equations of the form \( ax^2 + bx + c = 0 \). When dealing with equations that can't be readily factored, this formula comes to the rescue:\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]In the context of our hyperbolic expression, the equation \( e^{2y} - 2xe^{y} + 1 = 0 \) is formed. Here, we identify:

• \(a = 1\),
• \(b = -2x\),
• \(c = 1\)

Using the quadratic formula, we plug these values into the formula to isolate \( e^{y} \): \[e^{y} = \frac{2x \pm \sqrt{4x^2 - 4}}{2}\]By calculating this, we discover two potential solutions, but when selecting the branch that corresponds to real-world positive values, we use \( e^{y} = x + \sqrt{x^2 - 1} \). This step is pivotal in simplifying equations and ensuring solutions are real and applicable. Knowing how and when to apply the quadratic formula is essential for dealing with any polynomial equations that intersect throughout mathematics and physics.

Natural Logarithm

The natural logarithm, denoted as \( \ln(x) \), is a logarithm to the base \( e \), where \( e \approx 2.71828 \). In mathematics, it plays a crucial role in solving equations where an exponent is unknown. It is incredibly useful for inverse operations involving exponential functions.

Once we have \( e^{y} \) from the quadratic equation solved previously, the next step is to determine \( y \). By taking the natural logarithm on both sides:\[y = \ln(x + \sqrt{x^2 - 1})\]This operation effectively 'undoes' the exponent and isolates \( y \), which aligns perfectly with our hyperbolic inverse formula. Often, natural logarithms are used in fields such as calculus, physics, and complex system modeling.

• They simplify multiplication into addition, which is useful in many mathematical transformations.
• The natural logarithm is instrumental in solving growth and decay problems, especially where change rates are proportional to the current state.

This technique demonstrates its importance in the realm of inverse hyperbolic calculations, confirming many textbook expressions like \( \cosh^{-1}(x) = \ln(x + \sqrt{x^2 - 1}) \).